Advertisements
Advertisements
प्रश्न
Express the following as the product of sine and cosine.
cos 2A + cos 4A
बेरीज
उत्तर
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
shaalaa.com
Transformation Formulae
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Trigonometry - Exercise 4.3 [पृष्ठ ८८]
APPEARS IN
संबंधित प्रश्न
Prove that:
\[2\cos\frac{5\pi}{12}\cos\frac{\pi}{12} = \frac{1}{2}\]
Prove that:
cos 20° cos 40° cos 80° = \[\frac{1}{8}\]
Prove that:
sin 10° sin 50° sin 60° sin 70° = \[\frac{\sqrt{3}}{16}\]
Show that:
sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0
Prove that:
sin 23° + sin 37° = cos 7°
Prove that:
\[\sin\frac{x}{2}\sin\frac{7x}{2} + \sin\frac{3x}{2}\sin\frac{11x}{2} = \sin 2x \sin 5x .\]
Prove that:
\[\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A} = \cot 8A\]
Prove that:
\[\sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma) = 4 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\beta + \gamma}{2} \right) \sin \left( \frac{\gamma + \alpha}{2} \right)\]
If cosec A + sec A = cosec B + sec B, prove that tan A tan B = \[\cot\frac{A + B}{2}\].
If A, B, C are in A.P., then \[\frac{\sin A - \sin C}{\cos C - \cos A}\]=
