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the value of (sinq-cosq)^2+(sinq+cosq)^2 is equivalent to
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sin²q+cos²q+sin²q+2sinqcosq +cos²q 2sin²q+2sinqcosq+2cos²q 2(sin²q+sinqcosq+cos²q)
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pls explain please
For my answer, I simply factored out the equation you posted. But xen claims it's a trig identity, and that could be so as well.. That's all I can do.
(sinq-cosq)^2=\[\sin^{q}+\cos^{2}q-2sinqcosq\],also(sinq+cosq)^2=\[ \sin^{2}q+\cos^{2}q+2sinqcosq\] using formulas of (a+b)^2 and (a-b)^2 and the standard definition \[\sin^{2}q+\cos^{2}q=1\].... you get 1-2sinqcosq and 1+2sinqcosq now add both 1+1=2
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