Question

Prove that sinx+sin3x+sin5x+sin7x=4 cosx cos2x sin4x

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

sinx + sin3x + sin5x + sin7x

= (sinx + sin3x) + (sin5x + sin7x)

Formula :-

$\tt{\rightarrow sinC+sinD = 2sin\dfrac{C+D}{2}.cos\dfrac{C-D}{2}}$

Using this formula :-

$\tt{\rightarrow 2sin\dfrac{x+3x}{2}.cos\dfrac{3x-x}{2} + 2sin\dfrac{5x+7x}{2}.cos\dfrac{7x-5x}{2}}$

= 2sin2x.cosx + 2sin6x.cosx

= 2cosx [sin2x + sin6x ]

$\tt{\rightarrow 2cosx(2sin\dfrac{2x+6x}{2}.cos\dfrac{6x-2x}{2}}$

= 2cosx. (2sin4x.cos2x)

= 4cosx.cos2x.sin4x

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Some additional identities :-

★tan θ = sin θ /cos θ

★cot θ = cos θ / sin θ

★(sin² θ) + (cos² θ) = 1

★1 + tan² θ = sec² θ

★1 + cot² θ = cosec² θ

Answer 2

Answer:

→ sinx + sin3x + sin5x + sin7x

= (sinx + sin3x) + (sin5x + sin7x)

Formula:

$\tt{\rightarrow sinC+sinD = 2sin\dfrac{C+D}{2}.cos\dfrac{C-D}{2}}$

Using this formula:

$\tt{\rightarrow 2sin\dfrac{x+3x}{2}.cos\dfrac{3x-x}{2} + 2sin\dfrac{5x+7x}{2}.cos\dfrac{7x-5x}{2}}$

= 2sin2x.cosx + 2sin6x.cosx

= 2cosx [sin2x + sin6x ]

$\tt{\rightarrow 2cosx(2sin\dfrac{2x+6x}{2}.cos\dfrac{6x-2x}{2})}$

= 2cosx. (2sin4x.cos2x)

= 4cosx.cos2x.sin4x