Trigonometric Equation Solving via Identities

Solve an equation involving trigonometric functions by applying addition, double angle, or product-to-sum identities to find unknown angle values or trigonometric ratios.

gaokao 2015 Q13 View
13. Given $\sin \alpha + 2 \cos \alpha = 0$, the value of $2 \sin \alpha \cos \alpha - \cos ^ { 2 } \alpha$ is \_\_\_\_.
gaokao 2018 Q15 5 marks View
Given $\sin \alpha + \cos \beta = 1 , \cos \alpha + \sin \beta = 0$, then $\sin ( \alpha + \beta ) = \_\_\_\_$.
gaokao 2019 Q11 View
11. Given $a \in \left( 0 , \frac { \pi } { 2 } \right) , 2 \sin 2 \alpha = \cos 2 \alpha + 1$, then $\sin \alpha =$
A. $\frac { 1 } { 5 }$
B. $\frac { \sqrt { 5 } } { 5 }$
C. $\frac { \sqrt { 3 } } { 3 }$
D. $\frac { 2 \sqrt { 5 } } { 5 }$
gaokao 2019 Q10 View
10. Given $\alpha \in \left( 0 , \frac { \pi } { 2 } \right) , 2 \sin 2 \alpha = \cos 2 \alpha + 1$, then $\sin \alpha =$
A. $\frac { 1 } { 5 }$
B. $\frac { \sqrt { 5 } } { 5 }$
C. $\frac { \sqrt { 3 } } { 3 }$
D. $\frac { 2 \sqrt { 5 } } { 5 }$
gaokao 2021 Q11 View
11. If $\alpha \in \left( 0 , \frac { \pi } { 2 } \right) , \tan 2 \alpha = \frac { \cos \alpha } { 2 - \sin \alpha }$, then $\tan \alpha =$
A. $\frac { \sqrt { 15 } } { 15 }$
B. $\frac { \sqrt { 5 } } { 5 }$
C. $\frac { \sqrt { 5 } } { 3 }$
D. $\frac { \sqrt { 15 } } { 3 }$
gaokao 2021 Q9 View
9. If $a \in \left(0, \frac{\pi}{2}\right)$, $\tan 2a = \frac{\cos a}{2 - \sin a}$, then $\tan a =$
A. $\frac{\sqrt{15}}{15}$
B. $\frac{\sqrt{5}}{5}$
C. $\frac{\sqrt{5}}{3}$
D. $\frac{\sqrt{15}}{3}$
isi-entrance 2022 Q1 View
Suppose, for some $\theta \in \left[ 0 , \frac { \pi } { 2 } \right] , \frac { \cos 3 \theta } { \cos \theta } = \frac { 1 } { 3 }$. Then $( \cot 3 \theta ) \tan \theta$ equals
(A) $\frac { 1 } { 2 }$
(B) $\frac { 1 } { 3 }$
(C) $\frac { 1 } { 8 }$
(D) $\frac { 1 } { 7 }$
isi-entrance 2026 QB8 View
Let $\frac { \tan ( \alpha - \beta + \gamma ) } { \tan ( \alpha + \beta - \gamma ) } = \frac { \tan \beta } { \tan \gamma }$. Then
(A) $\sin ( \beta - \gamma ) = \sin ( \alpha - \beta )$.
(B) $\sin ( \alpha - \gamma ) = \sin ( \beta - \gamma )$.
(C) $\sin ( \beta - \gamma ) = 0$.
(D) $\sin 2 \alpha + \sin 2 \beta + \sin 2 \gamma = 0$
jee-advanced 2018 Q13 View
Let $a , b , c$ be three non-zero real numbers such that the equation $$\sqrt { 3 } a \cos x + 2 b \sin x = c , x \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right]$$ has two distinct real roots $\alpha$ and $\beta$ with $\alpha + \beta = \frac { \pi } { 3 }$. Then, the value of $\frac { b } { a }$ is $\_\_\_\_$.
jee-main 2012 Q67 View
Suppose $\theta$ and $\phi ( \neq 0 )$ are such that $\sec ( \theta + \phi )$, $\sec \theta$ and $\sec ( \theta - \phi )$ are in A.P. If $\cos \theta = k \cos \left( \frac { \phi } { 2 } \right)$ for some $k$, then $k$ is equal to
(1) $\pm \sqrt { 2 }$
(2) $\pm 1$
(3) $\pm \frac { 1 } { \sqrt { 2 } }$
(4) $\pm 2$
jee-main 2021 Q63 View
If $e ^ { \cos ^ { 2 } x + \cos ^ { 4 } x + \cos ^ { 6 } x + \ldots \infty \log _ { e } 2 }$ satisfies the equation $t ^ { 2 } - 9 t + 8 = 0$, then the value of $\frac { 2 \sin x } { \sin x + \sqrt { 3 } \cos x }$, where $0 < x < \frac { \pi } { 2 }$, is equal to
(1) $\frac { 3 } { 2 }$
(2) $\frac { 1 } { 2 }$
(3) $\sqrt { 3 }$
(4) $2 \sqrt { 3 }$
jee-main 2021 Q64 View
If $0 < x , y < \pi$ and $\cos x + \cos y - \cos ( x + y ) = \frac { 3 } { 2 }$, then $\sin x + \cos y$ is equal to:
(1) $\frac { 1 } { 2 }$
(2) $\frac { \sqrt { 3 } } { 2 }$
(3) $\frac { 1 - \sqrt { 3 } } { 2 }$
(4) $\frac { 1 + \sqrt { 3 } } { 2 }$
turkey-yks 2013 Q25 View
Given that $\alpha , \beta \in \left[ 0 , \frac { \pi } { 2 } \right]$,
$$\sin ( \alpha - \beta ) = \sin \alpha \cdot \cos \beta$$
Which of the following is true?
A) $\alpha = 0$ or $\beta = \frac { \pi } { 2 }$
B) $\alpha = 0$ or $\beta = \frac { \pi } { 4 }$
C) $\alpha = \frac { \pi } { 2 }$ or $\beta = 0$
D) $\alpha = \frac { \pi } { 2 }$ or $\beta = \frac { \pi } { 2 }$
E) $\alpha = \frac { \pi } { 4 }$ or $\beta = 0$