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Simplify: (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2
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a. 1 b. 2 c. sin^2Θ d. cos^2Θ
sin^2(x) + cos^2(x) -cosx*sin(x) +sin^2(x) + cos^2(x) +cos(x)*sin(x)= 2*sin^2(x) + 2 *cos^2(x)= 2
Did you trying distributing? Also use: \(\Large sin^2 \theta +cos^2\theta=1\)
you have to know sin^2(x) + cos^2(x) = 1
@Luigi0210 I got the second expression to be 1, but idk about the 1st expression
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Distribute it: \(\Large (sin\theta-cos\theta)^2=sin^2\theta-2cos\theta sin\theta+cos^2\theta\)
Like @borak said, it'll end up being 2 There will be \(\Large sin^2\theta+cos^2\theta+sin^2\theta+cos^2\theta+2cos\theta sin\theta-2cos\theta sin\theta\)
The sin's and cos's squared will add up, and the 2cos sin will cancel when distributed
so it's B (2)?
Yup yup ~
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thanks :)
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