How do you verify the identity ${(sin θ + cos θ)}^{2} - 1 = sin 2 θ$ $(sintheta+costheta)^2-1=sin2theta$ ?

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How do you verify the identity ${(sin θ + cos θ)}^{2} - 1 = sin 2 θ$ $(sintheta+costheta)^2-1=sin2theta$ ?

2 Answers

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Answer 1 · 2017-01-22T07:47:40

See proof below

Explanation:

We need

${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$

$sin 2 θ = 2 sin θ cos θ$ $sin2theta=2sinthetacostheta$

Therefore,

$L H S = {(cos θ + sin θ)}^{2} - 1$ $LHS=(costheta+sintheta)^2-1$

$= {cos}^{2} θ + {sin}^{2} θ + 2 sin θ cos θ - 1$ $=cos^2theta+sin^2theta+2sinthetacostheta-1$

$= 1 + 2 sin θ cos θ - 1$ $=1+2sinthetacostheta-1$

$= 2 sin θ cos θ$ $=2sinthetacostheta$

$= sin 2 θ$ $=sin2theta$

$= R H S$ $=RHS$

$Q E D$ $QED$

Answer 2 · 2017-01-22T07:48:01

$cancel1+2 sinthetacostheta cancel-1=2sinthetacostheta$
and the identity is verified.

Explanation:

Since

$sin2theta=2sinthetacostheta$ ,

you can substitute it in the given expression and expand the square of the bynomial:

$color(red)(sin^2theta+cos^2theta+2sinthetacostheta)-1 =color(green)(2 sinthetacostheta)$

Then, since

$sin^2theta+cos^2theta=1$ , you get:

$cancel1+2 sinthetacostheta cancel-1=2sinthetacostheta$

and the identity is verified.

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How do you verify the identity ${(sin θ + cos θ)}^{2} - 1 = sin 2 θ$ $(sintheta+costheta)^2-1=sin2theta$ ?

2 Answers

Explanation:

Explanation:

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How do you verify the identity (sintheta+costheta)^2-1=sin2theta(sinθ+cosθ)2−1=sin2θ(sintheta+costheta)^2-1=sin2theta?

2 Answers

Explanation:

Explanation:

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How do you verify the identity ${(sin θ + cos θ)}^{2} - 1 = sin 2 θ$ $(sintheta+costheta)^2-1=sin2theta$ ?