Class 11 Commerce Maths - Trigonometric Functions

$\sin \theta =\cos \theta$ for all values of $\theta$
Enter $1$ for true and $0$ for false

$\sin { { 30 }^{ \circ }=\frac { 1 }{ 2 } } ,\cos { { 30 }^{ \circ }=\frac { \sqrt { 3 } }{ 2 } } \\ \therefore \sin { \theta \neq \cos { \theta } }$
for all values of

$\theta$ .

$\cos{(A-B)}=$

From Trignnometric equations,

$\cos(A-B)=\cos A\cos B+\sin A\sin B$

Convert ${ 25 }^{ 0 }$ into Radian measure.

We know that

${ \pi }^{ c }={ 180 }^{ 0 }\\ \Longrightarrow { 1 }^{ 0 }={ \left( \frac { \pi }{ 180 } \right) }^{ c }\\ \Longrightarrow { 25 }^{ 0 }={ \left( \frac { 5\pi }{ 36 } \right) }^{ c }\\$

If $7\sin ^{ 2 }{ x } +3\cos ^{ 2 }{ x } =4$ , show that $\tan { x } =\dfrac { 1 }{ \sqrt { 3 } }$

$7\sin^{2}x+3\cos^{2}x=4$

$4\sin^{2}x+3\sin^{2}x+3\cos^{2}x=4$

$4\sin^{2}x+3=4$

$4\sin^{2}x=1$

$\sin^{2}x=\dfrac{1}{4}$

$\sin x=\dfrac{1}{2}$ or

$\sin x=-\dfrac{1}{2}$

Taking the positive root

$x=\dfrac{\pi}{6}$

$\tan (\dfrac{\pi}{6})=\dfrac{1}{\sqrt{3}}$

Hence proved.

Solve
$\cos (A - B) - \cos (180 - (A + B)=0$

$\cos (A-B)-\cos (180-(A+B))=-2\sin \bigg(\dfrac{A-B+180-A-B}{2}\bigg)\sin \bigg(\dfrac{A-B-180+A+B}{2}\bigg)$

$=-2\sin (90-B)\sin (A-90)=2\cos A\cos B$

$\cos (A-B)-\cos(180-(A+B))=0\implies \cos A=0$ or

$\cos B=0\implies A=90$ or

$B=90$

The value of $\cos \theta$ increases as $\theta$ increases.
Enter $1$ for true and $0$ for false

$\cos { { 45 }^{ \circ }=\frac { 1 }{ \sqrt { 2 } } },\cos { { 90 }^{ \circ }=0 }$
Here, as

$\theta$ increases from 45 to 90, value of

$cos\theta$ decreases from

$1/\sqrt2$ to 0. Thus, given statement is not correct.

Find the value of $\theta$ laying between 0 and $\displaystyle \frac{\pi}{2}$ and satisfying the equation

$\begin{vmatrix}1 + \cos^2 \theta & \sin^2 \theta & 4 \sin 4 \theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \sin 4 \theta\\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \sin 4 \theta\end{vmatrix} = 0$ .

If the value of $\theta = \displaystyle \frac{a \pi}{b}$ , then find the value of $\dfrac{(b-3)}a$

$\begin{vmatrix}1 + \cos^2 \theta & \sin^2 \theta & 4 \sin 4 \theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4\sin 4 \theta\\ \cos^2 \theta & \sin^2 \theta & 1 + 4\sin 4 \theta\end{vmatrix} = 0$

${ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }$

$\begin{vmatrix} 1 & -1 & 0 \\ cos^{ 2 }\theta & 1+sin^{ 2 }\theta & 4sin4\theta \\ 0 & -1 & 1 \end{vmatrix}=0$

${ C }_{ 2 }\rightarrow { C }_{ 2 }+{ C }_{ 1 }$

$\begin{vmatrix} 1 & 0 & 0 \\ cos^{ 2 }\theta & 2 & 4sin4\theta \\ 0 & -1 & 1 \end{vmatrix}=0$

$\Rightarrow 2+4\sin\theta=0$

$\Rightarrow \sin4\theta=-\dfrac{1}{2}$

$\Rightarrow \sin4\theta=-\sin (\dfrac{\pi}{6})$

$\Rightarrow \sin4\theta=\sin (\pi+\dfrac{\pi}{6})$

$\Rightarrow 4\theta=\dfrac{7\pi}{6}$

$\Rightarrow \theta=\dfrac{7\pi}{24}$

Comparing with the given value of

$\theta$ , we get

$a=7,b=24$

Value of

$\dfrac{b-3}{a}=3$

If $\Delta = \begin{vmatrix}1 & sin \theta & 1\\ - sin \theta & 1 & sin \theta\\ -1 & -sin \theta & 1\end{vmatrix} ; 0 \leq \theta < 2 \pi$ then $\Delta \in [a,b]$ Find $\dfrac{b}{a}?$

$\Delta = \begin{vmatrix}1 & sin \theta & 1\\ - sin \theta & 1 & sin \theta\\ -1 & -sin \theta & 1\end{vmatrix}$

$\Rightarrow \Delta =(1+\sin^2 \theta)-\sin\theta(-\sin\theta+\sin\theta)+(\sin^2\theta+1)$

$\Rightarrow \Delta=2+2\sin^2\theta$

$\sin^2\theta \in [0,1]$

$\Rightarrow \Delta \in [2,4]$

$\Rightarrow a=2$ and

$b=4$

$\therefore \displaystyle\frac{b}{a}=2$

The general solution of $\displaystyle 4\tan^{2}\theta=3\sec^{2}\theta$ is $\theta = n\pi\pm\dfrac{\pi}{m}$ . Then, find the value of $m$ .

$\displaystyle \because 4\tan^{2}\theta=3\sec^{2}\theta$

$\displaystyle \frac{4\sin^{2}\theta}{\cos^{2}\theta}=\frac{3}{\cos^{2}\theta}$

$\cos^{2}\theta\neq 0 \therefore \theta\neq \left ( 2n+1 \right )\frac{\pi }{2},n\in I$

Hence,

$4\sin^{2}\theta=3$

$\Rightarrow \sin^{2}\theta=\left ( \frac{\sqrt{3}}{2} \right )^{2}$

$\Rightarrow \sin^{2}\theta=\sin^{2}\frac{\pi }{3}$

$\Rightarrow \theta=n\pi \pm \frac{\pi }{3},n\in I$

Solve $\displaystyle \sin5x.\cos3x=\sin6x.\cos2x$ . then $\displaystyle \frac{n\pi }{2},n\in I$ or $\displaystyle n\pi \pm \frac{\pi }{6},n\in I$
If true then enter $1$ and if false then enter $0$

$\displaystyle \because \sin5x.\cos3x=\sin6x.\cos2x$

$\Rightarrow 2\sin5x.\cos3x=2\sin6x.\cos2x$

$\Rightarrow \sin8x+\sin2x=\sin8x+\sin4x$

$\Rightarrow \sin4x-\sin2x=0$

$\Rightarrow 2\sin2x.\cos2x-\sin2x=0$

$\Rightarrow \sin2x\left ( 2\cos2x-1 \right )=0$

$\Rightarrow \sin2x=0$ or

$\displaystyle 2 cos 2x-1=0$

$\Rightarrow 2x=n\pi \in I$ or

$\displaystyle \cos2x=\frac{1}{2}$

$\Rightarrow x=\frac{n\pi }{2},n\in I$

$\Rightarrow x=n\pi \pm \frac{\pi }{6},n\pi I$

$\therefore$ Solution of given equation is

$\displaystyle \frac{n\pi }{2},n\in I$ or

$\displaystyle n\pi \pm \frac{\pi }{6},n\in I$

Solve $\displaystyle \sin x+\cos x=\sqrt{2}$ , then $x=\displaystyle 2n\pi +\frac{\pi }{4},n\in I$
If true then enter $1$ and if false then enter $0$

$\displaystyle \because \sin x +\cos x=\sqrt{2}$

$\displaystyle \therefore$ divide both sides of equation (i) by

$\displaystyle \sqrt{2},$ we get,

$\displaystyle \sin x \frac{1}{\sqrt{2}}+\cos x \frac{1}{\sqrt{2}}=1$

$\Rightarrow \sin x\sin\frac{\pi }{4}+\cos x.\cos\frac{\pi}{4}=1$

$\Rightarrow \cos\left ( x-\frac{\pi }{4} \right )=1$ (Since,

$\sin A \sin B + \cos A \cos B =\cos (A- B)$ )

$\Rightarrow x-\frac{\pi }{4}=2n\pi ,n\in I$

$\Rightarrow x=2n\pi +\frac{\pi }{4},n\in I$

$\therefore$ Solution of given equation is

$\displaystyle 2n\pi +\frac{\pi }{4},n\in I$

Solve $\displaystyle \sin x+\cos x=1+\sin x.\cos x$ , then $2n\pi \pm \frac{\pi }{4}$
If true then enter $1$ and if false then enter $0$

$\displaystyle \because \sin x\cos x=1\sin x.\cos x$
Let

$\displaystyle \sin x+\cos x=t\Rightarrow \sin^{2}x+\cos^{2}x+2\sin x.\cos x=t^{2}\Rightarrow \sin x.\cos s=\frac{t^{2}-1}{2}$
Now Put

$\displaystyle \sin x +cos x=t$ and

$\displaystyle \sin x.\cos x=\frac{t^{2}-1}{2}$ in (i), we get

$\displaystyle t=1+\frac{t^{2}-1}{2}\Rightarrow t^{2}2t+1=0\Rightarrow t=1\Rightarrow \sin x\cos x=1\because t=\sin x+\cos x\Rightarrow \sin x +\cos x=1$
divide both sides of equation (ii) by

$\displaystyle \sqrt{2},$ we get

$\displaystyle \Rightarrow \sin x\frac{1}{\sqrt{2}}+\cos x.\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \cos\left ( x-\frac{\pi }{4} \right )=\cos\frac{\pi }{4}\Rightarrow x-\frac{\pi }{4}\Rightarrow x-\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4}$

Find the general solution of the equation $4\cos^{2} x = 1$

The given equation is

$4\cos^{2}x = 1$

$\Rightarrow \cos^{2} x = \dfrac {1}{4}$

$\Rightarrow \cos^{2}x = \left (\dfrac {1}{2}\right )^{2}$

$\cos^{2} x = \left (\cos \dfrac {\pi}{3}\right )^{2}$

$\Rightarrow \cos^{2}x = \cos^{2}\dfrac {\pi}{3}$

$x = \dfrac {\pi}{3}$

$\therefore$ The general solution,

$x = n\pi \pm \dfrac {\pi}{3}$ . where

$n\epsilon I$ .

Find the general solution of $\cos x + \sin x = 1$

We have,

$\cos x + \sin x = 1$

$\Rightarrow \cfrac {1}{\sqrt {2}} \cos x + \cfrac {1}{\sqrt {2}} \sin x = \cfrac {1}{\sqrt {2}}$ ....[Dividing both sides by

$\sqrt {2}$ ]

$\Rightarrow \cos \cfrac {\pi}{4}\cos x + \sin \cfrac {\pi}{4}\sin x = \cfrac {1}{\sqrt {2}}$

$\Rightarrow \cos \left (x - \cfrac {\pi}{4}\right ) = \cos \cfrac {\pi}{4}$

$\therefore x - \cfrac {\pi}{4} = 2n\pi \pm \cfrac {\pi}{4}$ , where

$n \in I$

$\Rightarrow x = 2n\pi \pm \cfrac {\pi}{4} + \cfrac {\pi}{4}$ , where

$n \in I$

$\Rightarrow x = 2n \pi + \cfrac {\pi}{4} + \cfrac {\pi}{4}, 2n\pi - \cfrac {\pi}{4} + \cfrac {\pi}{4}$ , where

$n \in I$

$\Rightarrow x = 2n\pi + \cfrac {\pi}{2}, 2n \pi$ , where

$n \in I$

$\Rightarrow x = (4n + 1) \cfrac {\pi}{2}, 2n\pi$ , where

$n \in I$

This is the required general solution.

If $\alpha + \beta - \gamma = \pi$ , and $\sin^{2} \alpha + \sin^{2}\beta - \sin^{2}\gamma = \lambda \sin \alpha \sin \beta \cos \gamma$ , then write the value of $\lambda$ .

Identity

$\sin^{2}A-\sin^{2}B$

$=\sin\left(A+B\right)\sin\left(A-B\right)$

$=\left(\sin\left(A \right )+\sin\left(B\right)\right)\left(\sin\left(A \right )-\sin\left(B \right )\right)$

$=2\sin\left ( \dfrac{A+B}{2} \right )\cos\left ( \dfrac{A-B}{2} \right ).2\cos\left ( \dfrac{A+B}{2} \right )\sin\left ( \dfrac{A+B}{2} \right )$

$=2\sin\left ( \dfrac{A+B}{2} \right )\cos\left ( \dfrac{A+B}{2} \right ).2\sin\left ( \dfrac{A-B}{2} \right )\cos\left ( \dfrac{A-B}{2} \right )$

$=\sin\left ( A+B \right )\sin\left ( A-B \right )$

Given equation

$\sin^{2}\alpha +\left [ \sin^{2}\beta -\sin^{2}\gamma \right ]$

$=\sin^{2}\alpha + \sin\left( \beta-\gamma\right)\sin\left(\beta + \gamma\right)$

$=\sin^{2}\alpha+\sin\left(\pi - \alpha\right) \sin\left(\beta + \gamma\right)$

$=\sin^{2} \alpha + \sin \alpha \sin\left(\beta + \gamma\right)$

$=\sin \alpha \left[\sin \alpha +\sin\left(\beta + \gamma\right)\right]$

$=\sin \alpha\left [\sin \left(\pi-\left(\beta-\gamma\right)\right)+\sin\left(\beta+\gamma\right)\right]$

$=\sin\alpha\left [\sin\left(\beta-\gamma\right)+\sin\left(\beta +\gamma\right)\right]$

$=\sin \alpha\left[2\sin \beta \cos \gamma\right]$

$=2\sin \alpha \sin \beta \cos \gamma$

$\therefore \lambda=2$

Find the domain of definition of the following function:
$\displaystyle y \, = \, arc\cos \, \frac{2x \, + \, 1}{2 \, \sqrt{2x}}$

$y= \text{arc }\cos(\cfrac { 2x+1 }{ 2\sqrt { 2x } } )$

$-1\le \cfrac { 2x+1 }{ 2\sqrt { 2x } }\le 1$

$-2\sqrt { 2x } \le 2x+1,$

$\Longrightarrow 8x\le 4{x}^{2}+4x+1,$

$\Longrightarrow 4{x}^{2}-4x+1\ge 0,$

$\Longrightarrow {(2x-1)}^{2}\ge 0$

This verifies for all

$x.$

${(2x-1)}^{2}\le { (2\sqrt { 2x } ) }^{ 2 }$

$\Longrightarrow 4{x}^{2}+4x+1\le 8x,$

$\Longrightarrow{(2x-1)}^{2}\le 0,$

$\Longrightarrow 2x-1=0,$

$\Longrightarrow x=\cfrac{1}{2},$

$\therefore$ So, domain will be

$\cfrac{1}{2}.$

How to calculate $\sin { 37 }$

We know that

${37}^{\circ}<{90}^{\circ}$

Hence

${37}^{\circ}$ is acute.

From

$\triangle{ABC},\,\,\angle{C}$ is acute.

$\therefore\,\angle{B}={90}^{\circ}$

$\therefore\,\,\angle{A}+\angle{B}+\angle{C}={180}^{\circ}$

$\Rightarrow\,\angle{C}={180}^{\circ}-\left(\angle{A}+\angle{B}\right)$

$\Rightarrow\,\angle{C}={180}^{\circ}-\left({37}^{\circ}+{90}^{\circ}\right)$

$\Rightarrow\,\angle{C}={180}^{\circ}-{127}^{\circ}={53}^{\circ}$

Let

$AB=3$ units,

$BC=4$ units then

By Pythagoras theorem,

${AC}^{2}={AB}^{2}+{BC}^{2}$

${AC}^{2}={3}^{2}+{4}^{2}=9+16=25$

$\Rightarrow\,AC=5$ units

$\therefore\,\sin{{37}^{\circ}}=\dfrac{AB}{AC}=\dfrac{3}{5}$

1452824_862880_ans_0a32cf62826d4a33a3f0f121a8c397d4.PNG

Solve the following equation:
$\displaystyle \tan \, x \, + \, \tan \, 2x \, + \, \tan \, 3x \, = \, 0$

$tanx + tan2x +tan3x = 0...(1)$

we have

$tan 3x = tan (x+2x) = (tanx + tan2x)(1-tanx .tan2x)$

$\Rightarrow tanx + tan2x = tan3x - tanx . tan2x. tanx3x$

using the above realation equation (1) can written as

$\Rightarrow tan3x - tan x. tan 2x .tan3x + tan 3x = 0$

$\Rightarrow tan 3x (2-tanx . tan2x)= 0$

$\therefore tan 3x = 0$

$x = x\pi/ 3 ; x\epsilon$ integers.

and,

$2- tanx . tan2x = 0$

$\Rightarrow tanx . tan 2x = 2$

$\Rightarrow tanx \dfrac{2tanx}{1-tan^2x} = x$

$\Rightarrow tan^2x = 1-tan^2x$

$\Rightarrow 2tan^2 x = 1$

$\Rightarrow tanx = \pm1/\sqrt{2}$

$x = n\pi\pm tan^{-1}(\pm 1/\sqrt{2})$

1137948_887981_ans_c113e0b53ba04cf9afdce42b91b7a262.jpg

Prove that $\cfrac { \sin { 9x } +\sin { 7x } +\sin { 5x } +\sin { 3x } }{ \cos { 9x } +\cos { 7x } +\cos { 5x } +\cos { 3x } } =\tan { 6x }$

Formulas :

$1)\sin(C+D)=2\sin(\dfrac {C+D} {2})\cos(\dfrac {C-D} {2})$

$2)\cos(C+D)=2\cos(\dfrac {C+D} {2})\cos(\dfrac {C-D} {2})$

Rewriting the above equation ;

$\dfrac {(\sin9x+\sin3x) +(\sin7x+\sin5x)} {(\cos9x+\cos3x) +(\cos7x+\cos5x)}$

$\Rightarrow \dfrac {2\sin(6x)\cos(2x)+2\sin(6x)\cos(x)}{2\cos(6x)\cos(2x)+2\cos(6x)\cos(x)}$

$\Rightarrow \dfrac {\sin6x} {\cos6x} =\tan6x$

Prove that $\sin(A-B)=\sin A \cos B-\cos A \sin B$

We can prove this by using euler’s identity : $e^{ix} = cosx + i\ sinx$

therefore we have $e^{iA} = cosA + i\ sinA$ ,

and $e^{iB} = cosB + isinB$

now

$e^{iA}*e^{iB} = e^{i(A+B)}$

$\Rightarrow (cosA + i\ sinA)(cosB + i\ sinB) = (cos(A+B) + i\ sin(A+B))$

$\Rightarrow (cosA*cosB - sinA*sinB) + i(sinA*cosB + cosA*sinB) = (cos(A+B) + i\ sin(A+B))$

On equating the imaginary parts on both sides of the equation we get the required result

$i.e. sin(A+B) = sinA*cosB + cosA*sinB$

Find the general solution of the following questions
$\cos x= 2x$
$\sin 2x + \cos x=0$ .

$\sin2x+cosx=0$

$\because \quad \sin2x=2sinxcosx$

$\Rightarrow \quad 2sinxcosx+cosx=0$

$\Rightarrow \quad cosx\left( 2sinx+1 \right) =0$

$\swarrow$

$\searrow$

$cosx=0$

$2sinx=-1$

$sinx=-1/2$

General solution for

$cosx=0$

$x=\left( 2n+1 \right) \pi /2\quad ,\quad n\in z$

$\Rightarrow \quad x=2n\pi \pm \pi /2\quad ,\quad n\in z$

General solution for

$sinx=-1/2$

let

$sinx=siny\quad \rightarrow \left( 1 \right)$

$sinx=-1/2\quad \rightarrow \left( 2 \right)$

From

$(1)$ &

$(2)$

$siny=-1/2$

$siny=\sin\dfrac { 7\pi }{ 6 }$

$y=\dfrac { 7\pi }{ 6 }$

General solution

$=x=n\pi +{ \left( -1 \right) }^{ n }y\quad ,\quad n\in z$

put

$y=\dfrac { 7\pi }{ 6 }$

$x=n\pi +{ \left( -1 \right) }^{ n }\dfrac { 7\pi }{ 6 } \quad ,\quad n\in z$

Hence for

$cosx=0,\quad x=\left( 2n+1 \right) \pi /2\quad ,\quad n\in z$

for

$sinx=-1/2,\quad x=n\pi +{ \left( -1 \right) }^{ n }\dfrac { 7\pi }{ 6 } \quad ,\quad n\in z$

Find the general solution for each of the following equation:
$\sin 2x + \cos x = 0$

Given

$\sin 2x+\cos x=0$

$\implies 2\sin x\cos x+\cos x=0$

$\implies (2\sin x+1)\cos x=0$

$\implies (2\sin x+1)=0$ or

$\cos x=0$

$\implies \sin x=-\dfrac 12$ or

$\cos x=0$

$\implies x=\dfrac{7\pi}{6}+2n\pi, x=\dfrac{11\pi}{6}+2n\pi$ or

$x=\pm\dfrac{\pi}{2}+2n\pi$ (Since,

$\sin$ is negative in third and fourth quadrant)

Solve $\sin x+ \sqrt3 \cos x = \sqrt2$

Solution:-

$\sin{x} + \sqrt{3}\cos{x} = \sqrt{2}$

$\sqrt{3} \cos{x} = \sqrt{2} - \sin{x}$

Squaring both sides, we have

${\left( \sqrt{3} \cos{x} \right)}^{2} = {\left( \sqrt{2} - \sin{x} \right)}^{2}$

$3\cos^{2}{x} = 2 + \sin^{2}{x} - 2\sqrt{2}\sin{x}$

$\Rightarrow \; 3 \left( 1 - \sin^{2}{x} \right) = 2 + \sin^{2}{x} - 2 \sqrt{2} \sin{x} \; \left( \because \sin^{2}{x} + \cos^{2}{x} = 1 \right)$

$3 - 3 \sin^{2}{x} = 2 + \sin^{2}{x} - 2 \sqrt{2} \sin{x}$

$\Rightarrow 4 \sin^{2}{x} - 2\sqrt{2} \sin{x} - 1 = 0$

The above equation is in the quadratic form, i.e.

$a{x}^{2} + bx + c$ .

Now, by using quadratic formula, i.e.,

$x = \cfrac{ -b \pm \sqrt{{b}^{2} - 4ac}}{2a}$ , we have

$\sin{x} = \cfrac{- \left(-2 \sqrt{2} \right) \pm \sqrt{ {\left( -2 \sqrt{2} \right)}^{2} - \left( 4 \times 4 \times \left( -1 \right) \right)}}{2 \times 4}$

$\Rightarrow \; \sin{x} = \cfrac{2 \sqrt{2} \pm \sqrt{8 + 16}}{8}$

$\Rightarrow \; \sin{x} = \cfrac{\sqrt{2} \pm \sqrt{6}}{4}$

$\therefore \; x = \sin^{-1}{\left( \cfrac{\sqrt{2} + \sqrt{6}}{4} \right)} + 2n\pi \; \text{ Or } \; x = \sin^{-1}{\left( \cfrac{\sqrt{2} - \sqrt{6}}{4} \right)} + 2n\pi \; \text{ For all n} \in \text{integer}$

Prove that
If $2 \, \sin \, \left(\theta + \dfrac{\pi}{3} \right) = \cos \left(\theta - \dfrac{\pi}{6} \right) \, then , \, \tan \theta + \sqrt{3} = 0$

$\sin(A+B)=\sin A\cos B+\sin B\cos A$

$\cos(A-B)=\cos A\cos B-\sin A\sin B$

$2\sin(\theta+\dfrac\pi3)=\cos(\theta-\dfrac\pi6)$

$\rightarrow 2(\sin\theta \cos\dfrac\pi3+\cos\theta \sin\dfrac\pi3)=\cos\theta \cos\dfrac\pi6+\sin\theta \sin\dfrac\pi6$

$\rightarrow 2(\dfrac {\sin\theta}2+\dfrac{\sqrt3 \cos\theta}2)=\dfrac{\sqrt3 \cos\theta}2+\dfrac{\sin\theta}2$

$\rightarrow 2(\sin\theta+\sqrt3 \cos\theta)= \sqrt3 \cos\theta+\sin\theta$

$\rightarrow \sin\theta=-\sqrt3\cos\theta$

$\rightarrow \dfrac{\sin\theta}{\cos\theta}=-\sqrt3$

$\rightarrow \tan\theta+\sqrt3=0$

Prove: $\dfrac{1 + sec \, A}{sec \, A } = \dfrac{sin^2 A}{1 - cos \, A}$

$L.H.S. =\dfrac{1+secA}{secA}$

$=\dfrac{1+\dfrac{1}{cosA}}{\dfrac{1}{cosA}}$ ...............

$\sec{x}=\dfrac{1}{\cos{x}}$

$=\dfrac{\dfrac{1+cosA}{cosA}}{\dfrac{1}{cosA}}$

$=\dfrac{1+cosA}{1}$

Now multiply the numerator and denominator with

$1-cosA$

$=\dfrac{(1+cosA)(1-cosA)}{1-cosA}$

$=\dfrac{1-cos^{2}A}{1-cosA}$ ...........

$a^2-b^2=(a-b)(a+b)$

$=\dfrac{sin^{2}A}{1-cosA}$ ..............

$[sin²θ+cos²θ=1]$

$=R.H.S.$

If $\cos (65^0-A)\cos(25^0+B)- \sin(65^0-A) \sin( 25^0+B)= \sin (m+A-B)$ .Find $m$

$cos(65^\circ-A)cos(25^\circ+B)-sin(65^\circ-A)sin(25^\circ+B)\\=cos[(65^\circ-A)+(25^\circ+B)]\\=cos[90^\circ-(A-B)]\\=sin(A-B)\\now\>compare\>it\>with\>given\>value\>sin(m+A-B)\>we\>get\>m\>=\>0$

Convert the angles:
a) $4.4^{c}$
b) $-\frac{9\pi}{2}^{c}$
into degree.

(a)

$2\pi^{c}=360^{\circ}$

$\Rightarrow 1^{c}=(\cfrac{180}{\pi})^{\circ}$

$\Rightarrow 4.4^{c}=\cfrac{180\times 7}{22} \times 4.4$

$=252^{\circ}$

$\therefore 4.4^{c}=252^{\circ}$

(b)

$2\pi^{c}=360^{\circ}$

$\Rightarrow \pi^{c}=180^{\circ}$

$\Rightarrow \cfrac{-9\pi^{c}}{2}=(\cfrac{-9}{2}\times 180)^{\circ}$

$=810^{\circ}$

$\therefore \cfrac{-9\pi^{c}}{2}=-810^{\circ}$

$ifsec\theta +tan\theta =p\quad then\quad the\quad value\quad of\quad cosec\theta$

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Trigonometric Functions - Class 11 Commerce Maths - Extra Questions

$\sin \theta =\cos \theta$ for all values of $\theta$
Enter $1$ for true and $0$ for false

$\cos{(A-B)}=$

Convert ${ 25 }^{ 0 }$ into Radian measure.

If $7\sin ^{ 2 }{ x } +3\cos ^{ 2 }{ x } =4$ , show that $\tan { x } =\dfrac { 1 }{ \sqrt { 3 } }$

Solve
$\cos (A - B) - \cos (180 - (A + B)=0$

The value of $\cos \theta$ increases as $\theta$ increases.
Enter $1$ for true and $0$ for false

If $\Delta = \begin{vmatrix}1 & sin \theta & 1\\ - sin \theta & 1 & sin \theta\\ -1 & -sin \theta & 1\end{vmatrix} ; 0 \leq \theta < 2 \pi$ then $\Delta \in [a,b]$ Find $\dfrac{b}{a}?$

The general solution of $\displaystyle 4\tan^{2}\theta=3\sec^{2}\theta$ is $\theta = n\pi\pm\dfrac{\pi}{m}$ . Then, find the value of $m$ .

Solve $\displaystyle \sin5x.\cos3x=\sin6x.\cos2x$ . then $\displaystyle \frac{n\pi }{2},n\in I$ or $\displaystyle n\pi \pm \frac{\pi }{6},n\in I$
If true then enter $1$ and if false then enter $0$

Solve $\displaystyle \sin x+\cos x=\sqrt{2}$ , then $x=\displaystyle 2n\pi +\frac{\pi }{4},n\in I$
If true then enter $1$ and if false then enter $0$

Solve $\displaystyle \sin x+\cos x=1+\sin x.\cos x$ , then $2n\pi \pm \frac{\pi }{4}$
If true then enter $1$ and if false then enter $0$

Find the general solution of the equation $4\cos^{2} x = 1$

Find the general solution of $\cos x + \sin x = 1$

If $\alpha + \beta - \gamma = \pi$ , and $\sin^{2} \alpha + \sin^{2}\beta - \sin^{2}\gamma = \lambda \sin \alpha \sin \beta \cos \gamma$ , then write the value of $\lambda$ .

Find the domain of definition of the following function:
$\displaystyle y \, = \, arc\cos \, \frac{2x \, + \, 1}{2 \, \sqrt{2x}}$

How to calculate $\sin { 37 }$

Solve the following equation:
$\displaystyle \tan \, x \, + \, \tan \, 2x \, + \, \tan \, 3x \, = \, 0$

Prove that $\cfrac { \sin { 9x } +\sin { 7x } +\sin { 5x } +\sin { 3x } }{ \cos { 9x } +\cos { 7x } +\cos { 5x } +\cos { 3x } } =\tan { 6x }$

Prove that $\sin(A-B)=\sin A \cos B-\cos A \sin B$

Find the general solution of the following questions
$\cos x= 2x$
$\sin 2x + \cos x=0$ .

Find the general solution for each of the following equation:
$\sin 2x + \cos x = 0$

Solve $\sin x+ \sqrt3 \cos x = \sqrt2$

Prove that
If $2 \, \sin \, \left(\theta + \dfrac{\pi}{3} \right) = \cos \left(\theta - \dfrac{\pi}{6} \right) \, then , \, \tan \theta + \sqrt{3} = 0$

Prove: $\dfrac{1 + sec \, A}{sec \, A } = \dfrac{sin^2 A}{1 - cos \, A}$

If $\cos (65^0-A)\cos(25^0+B)- \sin(65^0-A) \sin( 25^0+B)= \sin (m+A-B)$ .Find $m$

Convert the angles:
a) $4.4^{c}$
b) $-\frac{9\pi}{2}^{c}$
into degree.

$ifsec\theta +tan\theta =p\quad then\quad the\quad value\quad of\quad cosec\theta$

Class 11 Commerce Maths Extra Questions

Trigonometric Functions - Class 11 Commerce Maths - Extra Questions

sinθ=cosθ\sin \theta =\cos \theta for all values of θ\thetaEnter 11 for true and 00 for false

cos(A−B)=\cos{(A-B)}=

Convert 250{ 25 }^{ 0 } into Radian measure.

If 7sin2x+3cos2x=47\sin ^{ 2 }{ x } +3\cos ^{ 2 }{ x } =4, show that tanx=1√3\tan { x } =\dfrac { 1 }{ \sqrt { 3 } }

Solvecos(A−B)−cos(180−(A+B)=0\cos (A - B) - \cos (180 - (A + B)=0

The value of cosθ\cos \theta increases as θ\theta increases.Enter 11 for true and 00 for false

If Δ=|1sinθ1−sinθ1sinθ−1−sinθ1|;0≤θ<2π \Delta = \begin{vmatrix}1 & sin \theta & 1\\ - sin \theta & 1 & sin \theta\\ -1 & -sin \theta & 1\end{vmatrix} ; 0 \leq \theta < 2 \pi then Δ∈[a,b]\Delta \in [a,b] Find ba?\dfrac{b}{a}?

The general solution of 4tan2θ=3sec2θ\displaystyle 4\tan^{2}\theta=3\sec^{2}\theta is θ=nπ±πm\theta = n\pi\pm\dfrac{\pi}{m}. Then, find the value of mm.

Solve sin5x.cos3x=sin6x.cos2x\displaystyle \sin5x.\cos3x=\sin6x.\cos2x. then nπ2,n∈I\displaystyle \frac{n\pi }{2},n\in I or nπ±π6,n∈I\displaystyle n\pi \pm \frac{\pi }{6},n\in IIf true then enter 11 and if false then enter 00

Solve sinx+cosx=√2\displaystyle \sin x+\cos x=\sqrt{2}, then x=2nπ+π4,n∈Ix=\displaystyle 2n\pi +\frac{\pi }{4},n\in IIf true then enter 11 and if false then enter 00

Solve sinx+cosx=1+sinx.cosx\displaystyle \sin x+\cos x=1+\sin x.\cos x, then 2nπ±π42n\pi \pm \frac{\pi }{4} If true then enter 11 and if false then enter 00

Find the general solution of the equation 4cos2x=14\cos^{2} x = 1

Find the general solution of cosx+sinx=1\cos x + \sin x = 1

If α+β−γ=π\alpha + \beta - \gamma = \pi, and sin2α+sin2β−sin2γ=λsinαsinβcosγ\sin^{2} \alpha + \sin^{2}\beta - \sin^{2}\gamma = \lambda \sin \alpha \sin \beta \cos \gamma, then write the value of λ\lambda.

Find the domain of definition of the following function:y=arccos2x+12√2x\displaystyle y \, = \, arc\cos \, \frac{2x \, + \, 1}{2 \, \sqrt{2x}}

How to calculate sin37\sin { 37 }

Solve the following equation:tanx+tan2x+tan3x=0\displaystyle \tan \, x \, + \, \tan \, 2x \, + \, \tan \, 3x \, = \, 0

Prove that sin9x+sin7x+sin5x+sin3xcos9x+cos7x+cos5x+cos3x=tan6x\cfrac { \sin { 9x } +\sin { 7x } +\sin { 5x } +\sin { 3x } }{ \cos { 9x } +\cos { 7x } +\cos { 5x } +\cos { 3x } } =\tan { 6x }

Prove that sin(A−B)=sinAcosB−cosAsinB\sin(A-B)=\sin A \cos B-\cos A \sin B

Find the general solution of the following questionscosx=2x \cos x= 2xsin2x+cosx=0 \sin 2x + \cos x=0.

Find the general solution for each of the following equation:sin2x+cosx=0\sin 2x + \cos x = 0

Solve sinx+√3cosx=√2 \sin x+ \sqrt3 \cos x = \sqrt2

Prove thatIf 2sin(θ+π3)=cos(θ−π6)then,tanθ+√3=02 \, \sin \, \left(\theta + \dfrac{\pi}{3} \right) = \cos \left(\theta - \dfrac{\pi}{6} \right) \, then , \, \tan \theta + \sqrt{3} = 0

Prove: 1+secAsecA=sin2A1−cosA\dfrac{1 + sec \, A}{sec \, A } = \dfrac{sin^2 A}{1 - cos \, A}

If \cos (65^0-A)\cos(25^0+B)- \sin(65^0-A) \sin( 25^0+B)= \sin (m+A-B)\cos (65^0-A)\cos(25^0+B)- \sin(65^0-A) \sin( 25^0+B)= \sin (m+A-B).Find mm

Convert the angles:a) 4.4^{c}4.4^{c}b) -\frac{9\pi}{2}^{c}-\frac{9\pi}{2}^{c}into degree.

ifsec\theta +tan\theta =p\quad then\quad the\quad value\quad of\quad cosec\theta ifsec\theta +tan\theta =p\quad then\quad the\quad value\quad of\quad cosec\theta

Class 11 Commerce Maths Extra Questions

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$\sin \theta =\cos \theta$ for all values of $\theta$
Enter $1$ for true and $0$ for false

$\cos{(A-B)}=$

Convert ${ 25 }^{ 0 }$ into Radian measure.

If $7\sin ^{ 2 }{ x } +3\cos ^{ 2 }{ x } =4$ , show that $\tan { x } =\dfrac { 1 }{ \sqrt { 3 } }$

Solve
$\cos (A - B) - \cos (180 - (A + B)=0$

The value of $\cos \theta$ increases as $\theta$ increases.
Enter $1$ for true and $0$ for false

If $\Delta = \begin{vmatrix}1 & sin \theta & 1\\ - sin \theta & 1 & sin \theta\\ -1 & -sin \theta & 1\end{vmatrix} ; 0 \leq \theta < 2 \pi$ then $\Delta \in [a,b]$ Find $\dfrac{b}{a}?$

The general solution of $\displaystyle 4\tan^{2}\theta=3\sec^{2}\theta$ is $\theta = n\pi\pm\dfrac{\pi}{m}$ . Then, find the value of $m$ .

Solve $\displaystyle \sin5x.\cos3x=\sin6x.\cos2x$ . then $\displaystyle \frac{n\pi }{2},n\in I$ or $\displaystyle n\pi \pm \frac{\pi }{6},n\in I$
If true then enter $1$ and if false then enter $0$

Solve $\displaystyle \sin x+\cos x=\sqrt{2}$ , then $x=\displaystyle 2n\pi +\frac{\pi }{4},n\in I$
If true then enter $1$ and if false then enter $0$

Solve $\displaystyle \sin x+\cos x=1+\sin x.\cos x$ , then $2n\pi \pm \frac{\pi }{4}$
If true then enter $1$ and if false then enter $0$

Find the general solution of the equation $4\cos^{2} x = 1$

Find the general solution of $\cos x + \sin x = 1$

If $\alpha + \beta - \gamma = \pi$ , and $\sin^{2} \alpha + \sin^{2}\beta - \sin^{2}\gamma = \lambda \sin \alpha \sin \beta \cos \gamma$ , then write the value of $\lambda$ .

Find the domain of definition of the following function:
$\displaystyle y \, = \, arc\cos \, \frac{2x \, + \, 1}{2 \, \sqrt{2x}}$

How to calculate $\sin { 37 }$

Solve the following equation:
$\displaystyle \tan \, x \, + \, \tan \, 2x \, + \, \tan \, 3x \, = \, 0$

Prove that $\cfrac { \sin { 9x } +\sin { 7x } +\sin { 5x } +\sin { 3x } }{ \cos { 9x } +\cos { 7x } +\cos { 5x } +\cos { 3x } } =\tan { 6x }$

Prove that $\sin(A-B)=\sin A \cos B-\cos A \sin B$

Find the general solution of the following questions
$\cos x= 2x$
$\sin 2x + \cos x=0$ .

Find the general solution for each of the following equation:
$\sin 2x + \cos x = 0$

Solve $\sin x+ \sqrt3 \cos x = \sqrt2$

Prove that
If $2 \, \sin \, \left(\theta + \dfrac{\pi}{3} \right) = \cos \left(\theta - \dfrac{\pi}{6} \right) \, then , \, \tan \theta + \sqrt{3} = 0$

Prove: $\dfrac{1 + sec \, A}{sec \, A } = \dfrac{sin^2 A}{1 - cos \, A}$

If $\cos (65^0-A)\cos(25^0+B)- \sin(65^0-A) \sin( 25^0+B)= \sin (m+A-B)$ .Find $m$

Convert the angles:
a) $4.4^{c}$
b) $-\frac{9\pi}{2}^{c}$
into degree.

$ifsec\theta +tan\theta =p\quad then\quad the\quad value\quad of\quad cosec\theta$