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Question Number 10825 by Saham last updated on 26/Feb/17

$$\mathrm{Given}\mathrm{that}$$$$\sin \left(\mathrm{x}\right)-\sin \left(\mathrm{y}\right)=\sin \left(\theta \right)$$$$\cos \left(\mathrm{x}\right)+\cos \left(\mathrm{y}\right)=\cos \left(\theta \right)$$$$\mathrm{Show}\mathrm{that}$$$$\cos (\mathrm{x}+\mathrm{y})=-\frac{1}{2}$$

Answered by mrW1 last updated on 26/Feb/17

$${(\sin \left(\mathrm{x}\right)-\sin \left(\mathrm{y}\right))}^2={\sin }^2\left(\theta \right)$$$${\sin }^2\left(x\right)-2\sin \left(x\right)\sin \left(y\right)+{\sin }^2\left(y\right)={\sin }^2\left(\theta \right)...\left(i\right)$$$$$$$${(\cos \left(\mathrm{x}\right)+\cos \left(\mathrm{y}\right))}^2={\cos }^2\left(\theta \right)$$$${\cos }^2\left(x\right)+2\cos \left(x\right)\cos \left(y\right)+{\cos }^2\left(y\right)={\cos }^2\left(\theta \right)...\left(ii\right)$$$$$$$$\left(i\right)+\left(ii\right):$$$${\sin }^2\left(x\right)+{\cos }^2\left(x\right)+2\cos \left(x\right)\cos \left(y\right)-2\sin \left(x\right)\sin \left(y\right)+{\sin }^2\left(y\right)+{\cos }^2\left(y\right)={\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)$$$$1+2[\cos \left(x\right)\cos \left(y\right)-\sin \left(x\right)\sin \left(y\right)]+1=1$$$$1+2\cos (x+y)+1=1$$$$\Rightarrow \cos (x+y)=-\frac{1}{2}$$

Commented by Saham last updated on 27/Feb/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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