If A - B - C - prove that cos 2 A -cos 2 B -cos 2 C -1-2 sin A sin B cos C

We have LHS = cos^2 A + cos^2 B cos^2 C = 1/2(1+cos2A)+1/2(1+cos 2B)+1/2(1+cos 2C) =1/2 +1/2(cos 2A + cos 2B cos 2C) =1/2 +1/2 ( 1 4 sin A sin B cos C) =1 2 ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Multiples of an Angle

If A + B + C ...

Question

If A + B + C = $π$ , prove that $c o s^{2} A + c o s^{2} B - c o s^{2} C = 1 - 2 s i n A s i n B c o s C$

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Solution

We have LHS = ${cos}^{2} A + {cos}^{2} B - {cos}^{2} C$

$= \frac{1}{2} (1 + cos2A) + \frac{1}{2} (1 + cos 2 B) + \frac{1}{2} (1 + cos 2 C)$

= $\frac{1}{2} + \frac{1}{2} (cos 2 A + cos 2 B - cos 2 C)$

= $\frac{1}{2} + \frac{1}{2} (1 - 4 sin A sin B cos C)$

= $1 - 2 sin A sin B cos C = RHS.$

$∴ {cos}^{2} A + {cos}^{2} B - {cos}^{2} C = 1 - 2 sin A sin B cos C$

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If A + B + C = π, prove that cos2A+cos2B−cos2C = 1−2sinAsinBcosC

We have LHS = cos2A+cos2B−cos2C =12(1+cos2A)+12(1+cos 2B)+12(1+cos 2C) =12+12(cos 2A+cos 2B−cos 2C) =12+12(1−4 sin A sin B cos C) =1−2 sin A sin B cos C=RHS. ∴cos2A+cos2B−cos2C=1−2 sin A sin B cos C

If A + B + C = $π$ , prove that $c o s^{2} A + c o s^{2} B - c o s^{2} C = 1 - 2 s i n A s i n B c o s C$