Prove that cos left( frac{pi}{4} + x right) + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

We use the trigonometric identity $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$.Let $A = \frac{\pi}{4} + x

Vedclass · Accepted Answer

Vedclass · Answer

We use the trigonometric identity $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$.
Let $A = \frac{\pi}{4} + x$ and $B = \frac{\pi}{4} - x$.
Then,$\frac{A+B}{2} = \frac{(\frac{\pi}{4} + x) + (\frac{\pi}{4} - x)}{2} = \frac{\frac{\pi}{2}}{2} = \frac{\pi}{4}$.
And,$\frac{A-B}{2} = \frac{(\frac{\pi}{4} + x) - (\frac{\pi}{4} - x)}{2} = \frac{2x}{2} = x$.
Substituting these into the identity:
$L.H.S. = 2 \cos \left( \frac{\pi}{4} \right) \cos x$.
Since $\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$,we have:
$L.H.S. = 2 \times \frac{1}{\sqrt{2}} \cos x = \sqrt{2} \cos x = R.H.S$.

Prove that $\cos \left( \frac{\pi}{4} + x \right) + \cos \left( \frac{\pi}{4} - x \right) = \sqrt{2} \cos x$.

Similar Questions

Let $\tan \alpha = \frac{a}{a+1}$ and $\tan \beta = \frac{1}{2a+1}$, then $\alpha + \beta$ is

If $A + B = \frac{\pi}{4},$ then $(1 + \tan A)(1 + \tan B) = $

If $\cos(\alpha + \beta) = \frac{4}{5}$ and $\sin(\alpha - \beta) = \frac{5}{13}$,where $0 \le \alpha, \beta \le \frac{\pi}{4}$,then $\tan 2\alpha = $

If $A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$,then $\tan A \cdot \tan B+\tan B \cdot \tan C+\tan C \cdot \tan A$ is equal to

If $0 < A < B < \frac{\pi}{4}$,$\cos (A+B) = \frac{11}{61}$ and $\sin (A-B) = \frac{24}{25}$,then $\sin 2A + \sin 2B = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module