P.o.t.W. #10 Solution

SOLUTIONMethod 1double angle identity for cosine: $$\cos 2\alpha = 2{\cos ^2}\alpha - 1$$substituting $$\theta$$ for $$2\alpha$$ gives: $$\cos \theta = 2{\cos ^2}\left( {\frac{\theta }{2}} \right) - 1\;$$from the cosine rule: $$\cos \theta = \frac{{{x^2} + {x^2} - {a^2}}}{{2xx}} = \frac{{2{x^2} - {a^2}}}{{2{x^2}}}$$substituting $$2{\cos ^2}\left( {\frac{\theta }{2}} \right) - 1$$ for $$\cos \theta$$, gives:$$2{\cos ^2}\left( {\frac{\theta }{2}} \right) - 1 = \frac{{2{x^2} - {a^2}}}{{2{x^2}}}$$$$2{\cos ^2}\left( {\frac{\theta }{2}} \right) = \frac{{2{x^2} - {a^2}}}{{2{x^2}}} + 1$$ $$= \frac{{2{x^2} - {a^2}}}{{2{x^2}}} + \frac{{2{x^2}}}{{2{x^2}}}$$ $$= \frac{{4{x^2} - {a^2}}}{{2{x^2}}}$$ $$= \frac{{4{x^2}}}{{2{x^2}}} - \frac{{{a^2}}}{{2{x^2}}}$$ $$= 2 - \frac{{{a^2}}}{{2{x^2}}}$$ $$= 1 - \frac{{{a^2}}}{{4{x^2}}}$$$$\cos \left( {\frac{\theta }{2}} \right) = \sqrt {1 - \frac{{{a^2}}}{{4{x^2}}}}$$ Q.E.D.Method 2Consider the shaded right triangle shown in figure.

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