Advertisements
Advertisements
Question
Express the following as the product of sine and cosine.
cos 2A + cos 4A
Advertisements
Solution
cos 2A + cos 4A = 2 cos`(("2A + 4A")/2) cos (("2A - 4A")/2)` ...`[∵ cos "C" + cos "D" = 2 cos (("C + D")/2) cos (("C - D")/2)]`
= 2 cos `((6"A")/2) cos ((6 - 2"A")/2)`
= 2 cos(3A) cos (-A) ...[∵ cos(-θ) = cos θ]
= 2 cos 3A cos A
APPEARS IN
RELATED QUESTIONS
Prove that:
\[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right) = \tan 3x\]
Prove that:
Prove that:
Prove that:
If \[m \sin\theta = n \sin\left( \theta + 2\alpha \right)\], prove that \[\tan\left( \theta + \alpha \right) \cot\alpha = \frac{m + n}{m - n}\]
Write the value of sin \[\frac{\pi}{12}\] sin \[\frac{5\pi}{12}\].
Write the value of \[\sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{3\pi}{10}\]
If sin 2A = λ sin 2B, then write the value of \[\frac{\lambda + 1}{\lambda - 1}\]
Prove that:
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
Prove that:
2 cos `pi/13` cos \[\frac{9\pi}{13} + \text{cos} \frac{3\pi}{13} + \text{cos} \frac{5\pi}{13}\] = 0
