How do you verify ${cos}^{2} u - {sin}^{2} u = 2 {cos}^{2} u - 1$ $cos ^2 u - sin ^2 u = 2 cos^2 u -1$ ?

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How do you verify ${cos}^{2} u - {sin}^{2} u = 2 {cos}^{2} u - 1$ $cos ^2 u - sin ^2 u = 2 cos^2 u -1$ ?

1 Answer

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Answer 1 · 2015-06-09T07:59:58

We know that ${sin}^{2} u + {cos}^{2} u = 1$ $sin^2u + cos^2 u =1$

Explanation:

LHS= - ( ${sin}^{2} u - {cos}^{2} u$ $sin^2u - cos^2 u$ )
= - { $(1 - {cos}^{2} u) - {cos}^{2} u$ $(1-cos^2u) - cos^2 u$ } [ since ${sin}^{2} u + {cos}^{2} u = 1$ $sin^2u + cos^2 u =1$ ]
= $2 {cos}^{2} u - 1$ $2cos^2u-1$
=RHS

Hence proved

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How do you verify ${cos}^{2} u - {sin}^{2} u = 2 {cos}^{2} u - 1$ $cos ^2 u - sin ^2 u = 2 cos^2 u -1$ ?

1 Answer

Explanation:

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How do you verify cos2u−sin2u=2cos2u−1cos ^2 u - sin ^2 u = 2 cos^2 u -1?

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How do you verify ${cos}^{2} u - {sin}^{2} u = 2 {cos}^{2} u - 1$ $cos ^2 u - sin ^2 u = 2 cos^2 u -1$ ?