If $\sin 135^\circ=\sqrt{1-y^2}$ and $\cos 60^\circ=\sqrt{1-x^2}$, find $\cos 67.5^\circ$ and $\sin 120^\circ$ in terms of $x$ and/or $y$

Question

Given that $\sin 135^\circ = \sqrt{1-y^2}$ and $\cos 60^\circ=\sqrt{1-x^2}$. Find in terms of $x$ and/or $y$:

• $\cos 67.5^\circ$
• $\sin 120^\circ$

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How do i relate this with half-angle formula?

Answer 1

Hint: Use that $$\sin(2x)=2\sin(x)\cos(x)$$ and $$\cos^2(2x)=2\cos^2(x)-1$$

Answer 2

$$(a)\ \sin 135^{\circ}=\sqrt{1-y^2}\Rightarrow y^2=1-\sin^2 135^{\circ}=\cos^2 135^{\circ}=\cos^2 (2 \cdot 67^{\circ})$$ $$\Rightarrow (2\cos^2 67^{\circ}-1)^2=y^2 \Rightarrow \cos 67^{\circ} = \sqrt{\frac{1\pm y}{2}}$$ $$(b)\ \cos 60^{\circ}=\sqrt{1-x^2}\Rightarrow x^2 = 1-\cos^2 60^{\circ}=\sin^2 60^{\circ}\Rightarrow \sin 60^{\circ}=\pm x$$ $$\sin 2\theta=2\sin \theta \cos \theta \Rightarrow \sin 120^{\circ}=\pm 2x\sqrt{1-x^2}$$