Question

Prove that:
. sin 120 - sin 40 = 2 cos 80.sin 40.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Question :

To prove

sin120°-sin40° = 2cos80°sin40°

Trignometric Formulas:

$1) \sin(c) + \sin(d) = 2 \sin( \frac{c + d}{2} ) \cos( \frac{c - d}{2} )$

$2) \sin(c) - \sin(d) = 2 \sin( \frac{c - d}{2} ) \cos( \frac{c + d}{2} )$

$3) \cos(c ) + \cos(d) = 2 \cos( \frac{c + d}{2} ) \cos( \frac{c - d}{2} )$

$4) \cos(c) - \cos(d) = - 2 \sin( \frac{c + d}{2} ) \sin( \frac{c - d}{2} )$

Solution :

We have to prove

sin120°-sin40° = 2cos80°sin40°

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Proof :

LHS =sin120°-sin40°

Apply sin c- sin d formula

$= 2 \sin( \frac{120 - 40}{2} ) \cos( \frac{120 + 40}{?} )$

$= 2 \sin(40) \degree \cos(80) \degree$

RHS= 2cos80°sin40°

⇒LHS = RHS

$\huge{\bold{ Hence\: Proved}}$

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More Trigonometry Formulas:

1. sin²A + cos²A = 1
2. sec²A - tan²A = 1
3. cosec²A - cot²A = 1
4. sin2A = 2 sinA cosA
5. cos2A = cos²A - sin²A
6. tan2A = 2 tanA / (1 - tan²A)