Question
Suppose that $a^{2}+b^{2}=1$ and $c^{2}+d^{2}=1 .$ Prove that $|a c+b d| \leq 1 .$ Hint: Let $a=\cos \theta, b=\sin \theta, c=\cos \phi$and $d=\sin \phi$
Step 1
This implies that $a$ and $b$ are the cosine and sine of some angle $\theta$, and $c$ and $d$ are the cosine and sine of some angle $\phi$. So, we can write $a=\cos \theta, b=\sin \theta, c=\cos \phi$ and $d=\sin \phi$. Show more…
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