Question Number 133929 by bemath last updated on 25/Feb/21
$$\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{76}°+\mathrm{cos}\:^{\mathrm{2}} \mathrm{16}°−\mathrm{cos}\:\mathrm{76}°.\mathrm{cos}\:\mathrm{16}°=? \\ $$
Answered by liberty last updated on 25/Feb/21
![⇒a^2 +b^2 =(a+b)^2 −2ab cos^2 76°+cos^2 16° −cos 76°.cos 16°= [ cos 76°+cos 16° ]^2 −3cos 76° cos 16° = [ 2cos 46° cos 30° ]^2 −(3/2) [ cos 92°+cos 60° ] =3cos^2 46°−(3/2)cos 92°−(3/4) =3[(1/2)+(1/2)cos 92° ]−(3/2)cos 92°−(3/4) = (3/2) − (3/4) = (3/4)](https://www.tinkutara.com/question/Q133930.png)
$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab} \\ $$$$ \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \mathrm{76}°+\mathrm{cos}\:^{\mathrm{2}} \mathrm{16}°\:−\mathrm{cos}\:\mathrm{76}°.\mathrm{cos}\:\mathrm{16}°= \\ $$$$\left[\:\mathrm{cos}\:\mathrm{76}°+\mathrm{cos}\:\mathrm{16}°\:\right]^{\mathrm{2}} −\mathrm{3cos}\:\mathrm{76}°\:\mathrm{cos}\:\mathrm{16}°\: \\ $$$$=\:\left[\:\mathrm{2cos}\:\mathrm{46}°\:\mathrm{cos}\:\mathrm{30}°\:\right]^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\:\left[\:\mathrm{cos}\:\mathrm{92}°+\mathrm{cos}\:\mathrm{60}°\:\right] \\ $$$$=\mathrm{3cos}\:^{\mathrm{2}} \mathrm{46}°−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:\mathrm{92}°−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$=\mathrm{3}\left[\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{92}°\:\right]−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:\mathrm{92}°−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}\:−\:\frac{\mathrm{3}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Answered by EDWIN88 last updated on 25/Feb/21
![cos^2 76°+cos^2 16°−cos 76°.cos 16° = (1/2) [ 1+cos 152°+1+cos 32°−cos 92°−cos 60° ] = (1/2) [ 2−(1/2)+cos 152°+cos 32°−cos 92° ] = (1/2) [ (3/2)+2cos 92°cos 60°−cos 92° ] = (1/2) [ (3/2)+cos 92°−cos 92° ]=(3/4)](https://www.tinkutara.com/question/Q133931.png)
$$\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{76}°+\mathrm{cos}\:^{\mathrm{2}} \mathrm{16}°−\mathrm{cos}\:\mathrm{76}°.\mathrm{cos}\:\mathrm{16}°\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\mathrm{1}+\mathrm{cos}\:\mathrm{152}°+\mathrm{1}+\mathrm{cos}\:\mathrm{32}°−\mathrm{cos}\:\mathrm{92}°−\mathrm{cos}\:\mathrm{60}°\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{152}°+\mathrm{cos}\:\mathrm{32}°−\mathrm{cos}\:\mathrm{92}°\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2cos}\:\mathrm{92}°\mathrm{cos}\:\mathrm{60}°−\mathrm{cos}\:\mathrm{92}°\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{92}°−\mathrm{cos}\:\mathrm{92}°\:\right]=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$