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Question Number 108085 by bemath last updated on 14/Aug/20

((⋓BeMath⋓)/∞) sin^8 75°−cos^8 75° =

$$\:\:\:\:\:\frac{\Cup\mathcal{B}{e}\mathcal{M}{ath}\Cup}{\infty} \\ $$$$\:\:\:\mathrm{sin}\:^{\mathrm{8}} \mathrm{75}°−\mathrm{cos}\:^{\mathrm{8}} \mathrm{75}°\:= \\ $$

Commented by bemath last updated on 14/Aug/20

thank you both

$${thank}\:{you}\:{both} \\ $$

Answered by Dwaipayan Shikari last updated on 14/Aug/20

(sin^4 θ+cos^4 θ)(sin^2 θ−cos^2 θ)(sin^2 θ+cos^2 θ) ((sin^2 θ+cos^2 θ)^2 −2cos^2 θsin^2 θ)(−cos2θ) (1−(1/2)sin^2 2θ)(−cos150°) (1−(1/2)((1/2))^2 )((√3)/2) (7/8).((√3)/2) =((7(√3))/(16))

$$ \\ $$$$\left({sin}^{\mathrm{4}} \theta+{cos}^{\mathrm{4}} \theta\right)\left({sin}^{\mathrm{2}} \theta−{cos}^{\mathrm{2}} \theta\right)\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right) \\ $$$$\left(\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} −\mathrm{2}{cos}^{\mathrm{2}} \theta{sin}^{\mathrm{2}} \theta\right)\left(−{cos}\mathrm{2}\theta\right) \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \mathrm{2}\theta\right)\left(−{cos}\mathrm{150}°\right) \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\frac{\mathrm{7}}{\mathrm{8}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{16}} \\ $$$$ \\ $$

Answered by john santu last updated on 14/Aug/20

a=sin 75° ; b=cos 75°=sin 15° { (J),(S) :} {: (★),(★) } ⇒J= a^8 −b^8 = (a^4 +b^4 )(a^4 −b^4 ) = (a^4 +b^4 )(a^2 +b^2 )(a^2 −b^2 ) = (a^4 +b^4 )(a^2 +b^2 )(a+b)(a−b) ⋇) a+b=sin 75°+sin 15°=2sin 45°cos 30° ⋇) a−b = sin 75°−sin 15°=2cos 45°sin 30° ⇒a+b=2×((√2)/2)×((√3)/2)=((√6)/2) ⇒a−b=2×((√2)/2)×(1/2)=((√2)/2) ⋇)a^2 +b^2 = 1 ⇒a^4 +b^4 = 1−2(ab)^2 =1−(1/2)(2ab)^2 =1−(1/8)=(7/8) ∗)2ab=2sin 75°cos 75°=sin 150°=(1/2) J= (7/8)×1×((√6)/2)×((√2)/2) = ((7×2(√3))/(32)) = ((14(√3))/(32))

$${a}=\mathrm{sin}\:\mathrm{75}°\:;\:{b}=\mathrm{cos}\:\mathrm{75}°=\mathrm{sin}\:\mathrm{15}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\begin{cases}{{J}}\\{{S}}\end{cases}\left.\begin{matrix}{\bigstar}\\{\bigstar}\end{matrix}\right\} \\ $$$$\Rightarrow{J}=\:{a}^{\mathrm{8}} −{b}^{\mathrm{8}} =\:\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right)\left({a}^{\mathrm{4}} −{b}^{\mathrm{4}} \right) \\ $$$$=\:\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$$=\:\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}+{b}\right)\left({a}−{b}\right) \\ $$$$\left.\divideontimes\right)\:{a}+{b}=\mathrm{sin}\:\mathrm{75}°+\mathrm{sin}\:\mathrm{15}°=\mathrm{2sin}\:\mathrm{45}°\mathrm{cos}\:\mathrm{30}° \\ $$$$\left.\divideontimes\right)\:{a}−{b}\:=\:\mathrm{sin}\:\mathrm{75}°−\mathrm{sin}\:\mathrm{15}°=\mathrm{2cos}\:\mathrm{45}°\mathrm{sin}\:\mathrm{30}° \\ $$$$\Rightarrow{a}+{b}=\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$$\Rightarrow{a}−{b}=\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\left.\divideontimes\right){a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\:\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\:\mathrm{1}−\mathrm{2}\left({ab}\right)^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{ab}\right)^{\mathrm{2}} \:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{7}}{\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left.\ast\right)\mathrm{2}{ab}=\mathrm{2sin}\:\mathrm{75}°\mathrm{cos}\:\mathrm{75}°=\mathrm{sin}\:\mathrm{150}°=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${J}=\:\frac{\mathrm{7}}{\mathrm{8}}×\mathrm{1}×\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\:\frac{\mathrm{7}×\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{32}}\:\:\:\:\:\:\:\: \\ $$$$=\:\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{32}} \\ $$

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