How do I prove the identities of these questions? 1+tan2u1−tan2u=1cos2u−sin2u sin⁡x+sin⁡(5x)cos⁡x+cos⁡(5x)=tan⁡(3x)

You will need two facts, first the definition of tangent, second main trigonometric identity: tan⁡x=sin⁡xcos⁡x sin2x+cos2x=1 Using that you may transform (1) as: 1+tan2u1−tan2u=1+sin2ucos2u1−sin2ucos2u= =cos2u+sin2ucos2ucos2u−sin2ucos2u=1cos2u⋅cos2ucos2u−sin2u=1cos2u−sin2u The original (2) which is: sin⁡x+sin⁡5xcos⁡x+cos⁡5x=tan⁡3x is false as shown in another answer (take x=π4 for instance). Here is a visualization of LHS and RHS

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trefoniu1

Answered question

2022-01-28

How do I prove the identities of these questions?

\frac{1 + \tan^{2} u}{1 - \tan^{2} u} = \frac{1}{\cos^{2} u - \sin^{2} u}

\frac{\sin x + \sin (5 x)}{\cos x + \cos (5 x)} = \tan (3 x)

Answer & Explanation

ul2ph3ojc

Beginner2022-01-29Added 12 answers

You will need two facts, first the definition of tangent, second main trigonometric identity:

\tan x = \frac{\sin x}{\cos x}

\sin^{2} x + \cos^{2} x = 1

Using that you may transform (1) as:

\frac{1 + \tan^{2} u}{1 - \tan^{2} u} = \frac{1 + \frac{\sin^{2} u}{\cos^{2} u}}{1 - \frac{\sin^{2} u}{\cos^{2} u}} =

= \frac{\frac{\cos^{2} u + \sin^{2} u}{\cos^{2} u}}{\frac{\cos^{2} u - \sin^{2} u}{\cos^{2} u}} = \frac{1}{\cos^{2} u} \cdot \frac{\cos^{2} u}{\cos^{2} u - \sin^{2} u} = \frac{1}{\cos^{2} u - \sin^{2} u}

The original (2) which is:

\sin x + \frac{\sin 5 x}{\cos x + \cos 5 x} = \tan 3 x

is false as shown in another answer (take

x = \frac{π}{4}

for instance). Here is a visualization of LHS and RHS

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