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What's the relationship between sine and cosine of complementary angles
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sin x = cos (90 - x) cos x = sin(90 - x)
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Here is a reminder. |dw:1468345595850:dw| Triangle ABC is a right triangle with right angle C. That means angles A and B must be complementary.
Using the trig ratios, we get: \(\sin A = \dfrac{y}{h}\) \(\cos B = \dfrac{y}{h}\) We can write \(\sin A = \cos B\) Since angles A and B are complementary, we have A + B = 90 or B = 90 - A, so we end up with \(\sin A = \cos (90 - A)\)
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