High School Trigonometry (Part 3)

Geometry Level 3

sin ⁡ 2 2 0 ∘ + cos ⁡ 2 5 0 ∘ + sin ⁡ 2 0 ∘ cos ⁡ 5 0 ∘ = ? \sin^{2}20^{\circ}+\cos^{2}50^{\circ}+\sin20^{\circ}\cos50^{\circ} =\textbf{?}


The answer is 0.75.

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Jessica Wang by Jessica Wang , 22, United Kingdom

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3 solutions

Potsawee Manakul
Jul 18, 2015

sin ⁡ 2 2 0 ∘ + cos ⁡ 2 5 0 ∘ + sin ⁡ 2 0 ∘ cos ⁡ 5 0 ∘ \sin^{2}20^{\circ}+\cos^{2}50^{\circ}+\sin20^{\circ}\cos50^{\circ} = ( sin ⁡ 2 0 ∘ + cos ⁡ 5 0 ∘ ) 2 − sin ⁡ 2 0 ∘ cos ⁡ 5 0 ∘ =(\sin20^{\circ}+\cos50^{\circ})^{2}-\sin20^{\circ}\cos50^{\circ} = ( sin ⁡ 2 0 ∘ + sin ⁡ 4 0 ∘ ) 2 − sin ⁡ 2 0 ∘ cos ⁡ 5 0 ∘ =(\sin20^{\circ}+\sin40^{\circ})^{2}-\sin20^{\circ}\cos50^{\circ} = ( 2 sin ⁡ 3 0 ∘ cos ⁡ 1 0 ∘ ) 2 − 1 2 ( sin ⁡ 7 0 ∘ − sin ⁡ 3 0 ∘ ) =(2\sin30^{\circ}\cos10^{\circ})^{2}-\frac{1}{2}(\sin70^{\circ}-\sin30^{\circ}) = cos ⁡ 2 1 0 ∘ − 1 2 ( sin ⁡ 7 0 ∘ − sin ⁡ 3 0 ∘ ) =\cos^{2}10^{\circ}-\frac{1}{2}(\sin70^{\circ}-\sin30^{\circ}) = cos ⁡ 2 0 ∘ + 1 2 − cos ⁡ 2 0 ∘ 2 + sin ⁡ 3 0 ∘ 2 = 0.75 =\frac{\cos20^{\circ}+1}{2}-\frac{\cos 20^{\circ}}{2}+\frac{\sin 30^{\circ}}{2} =0.75

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@Potsawee Manakul Same method, upvoted! x

Jessica Wang - 5 years, 11 months ago
Chew-Seong Cheong
Jul 18, 2015

sin ⁡ 2 2 0 ∘ + cos ⁡ 2 5 0 ∘ + sin ⁡ 2 0 ∘ cos ⁡ 5 0 ∘ = cos ⁡ 2 7 0 ∘ + cos ⁡ 2 5 0 ∘ + cos ⁡ 7 0 ∘ cos ⁡ 5 0 ∘ = ( cos ⁡ 7 0 ∘ + cos ⁡ 5 0 ∘ ) 2 − cos ⁡ 7 0 ∘ cos ⁡ 5 0 ∘ = ( cos ⁡ ( 60 + 10 ) ∘ + cos ⁡ ( 60 − 10 ) ∘ ) 2 − cos ⁡ ( 60 + 10 ) ∘ cos ⁡ ( 60 − 10 ) ∘ = ( 2 cos ⁡ 6 0 ∘ cos ⁡ 1 0 ∘ ) 2 − ( cos ⁡ 6 0 ∘ cos ⁡ 1 0 ∘ − sin ⁡ 6 0 ∘ sin ⁡ 1 0 ∘ ) ( cos ⁡ 6 0 ∘ cos ⁡ 1 0 ∘ + sin ⁡ 6 0 ∘ sin ⁡ 1 0 ∘ ) = cos ⁡ 2 1 0 ∘ − ( 1 2 cos ⁡ 1 0 ∘ − 3 2 sin ⁡ 1 0 ∘ ) ( 1 2 cos ⁡ 1 0 ∘ + 3 2 sin ⁡ 1 0 ∘ ) = cos ⁡ 2 1 0 ∘ − 1 4 cos ⁡ 2 1 0 ∘ + 3 4 sin ⁡ 2 1 0 ∘ = 3 4 cos ⁡ 2 1 0 ∘ + 3 4 sin ⁡ 2 1 0 ∘ = 3 4 = 0.75 \sin^2{20^\circ} + \cos^2{50^\circ} + \sin{20^\circ} \cos{50^\circ} \\ = \cos^2{70^\circ} + \cos^2{50^\circ} + \cos{70^\circ} \cos{50^\circ} \\ = (\cos{70^\circ} + \cos{50^\circ})^2 - \cos{70^\circ} \cos{50^\circ} \\ = (\cos{(60+10)^\circ} + \cos{(60-10)^\circ})^2 - \cos{(60+10)^\circ} \cos{(60-10)^\circ} \\ = (2\cos{60^\circ}\cos{10^\circ})^2 \\ \quad - (\cos{60^\circ}\cos{10^\circ} - \sin{60^\circ}\sin{10^\circ}) (\cos{60^\circ}\cos{10^\circ} + \sin{60^\circ}\sin{10^\circ}) \\ = \cos^2{10^\circ} - \left(\frac{1}{2}\cos{10^\circ} - \frac{\sqrt{3}}{2}\sin{10^\circ} \right) \left(\frac{1}{2}\cos{10^\circ} + \frac{\sqrt{3}}{2}\sin{10^\circ} \right) \\ = \cos^2{10^\circ} - \frac{1}{4}\cos^2{10^\circ} + \frac{3}{4}\sin^2{10^\circ} \\ = \frac{3}{4}\cos^2{10^\circ} + \frac{3}{4}\sin^2{10^\circ} \\ = \frac{3}{4} = \boxed{0.75}

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Great solution

Sai Ram - 5 years, 10 months ago
Raghav Chaudhary
Aug 6, 2015

Consider the identity a 3 − b 3 = ( a − b ) ( a 2 + b 2 + a b ) a^{3} -b^{3}=(a-b)(a^{2}+b^{2}+ab)

We have sin ⁡ 3 2 0 ∘ − cos ⁡ 3 5 0 ∘ = ( sin ⁡ 2 0 ∘ − cos ⁡ 5 0 ∘ ) ( sin ⁡ 2 2 0 ∘ + cos ⁡ 2 5 0 ∘ + s i n 2 0 ∘ c o s 5 0 ∘ ) \sin^{3}20^{\circ}-\cos^{3}50^{\circ} =(\sin20^{\circ}-\cos50^ {\circ})(\sin^{2}20^{\circ}+\cos^{2}50^{\circ}+sin20^{\circ}cos50^{\circ})

So, we now need to evaluate

sin ⁡ 3 2 0 ∘ − cos ⁡ 3 5 0 ∘ sin ⁡ 2 0 ∘ − cos ⁡ 5 0 ∘ \frac{\sin^{3}20^{\circ}-\cos^{3}50^{\circ}}{\sin20^{\circ}-\cos50^ {\circ}}

We now use the identity

sin ⁡ 3 θ = 3 sin ⁡ θ − 4 sin ⁡ 3 θ \sin3\theta =3\sin\theta - 4\sin^{3}\theta cos ⁡ 3 θ = 4 cos ⁡ 3 θ − 3 cos ⁡ θ \cos3\theta =4\cos^{3}\theta - 3\cos\theta

Now, sin ⁡ 6 0 ∘ = 3 sin ⁡ 2 0 ∘ − 4 sin ⁡ 3 2 0 ∘ . . . . . . . . . . . . . . ( 1 ) \sin60^{\circ} =3\sin20^{\circ} - 4\sin^{3}20^{\circ} .............. (1) cos ⁡ 15 0 ∘ = 4 cos ⁡ 3 5 0 ∘ − 3 cos ⁡ 2 0 ∘ . . . . . . . . . . . . . . ( 2 ) \cos150^{\circ} =4\cos^{3}50^{\circ} - 3\cos20^{\circ} .............. (2)

Realizing cos ⁡ 15 0 ∘ = − sin ⁡ 6 0 ∘ \cos150^{\circ} = - \sin60^{\circ} , we rewrite (2) as:

− sin ⁡ 6 0 ∘ = 4 cos ⁡ 3 5 0 ∘ − 3 cos ⁡ 2 0 ∘ . . . . . . . . . . . . . . ( 3 ) - \sin60^{\circ} = 4\cos^{3}50^{\circ} - 3\cos20^{\circ} .............. (3)

Adding (1) and (3), we have 0 = 3 ( sin ⁡ 2 0 ∘ − cos ⁡ 5 0 ∘ ) − 4 ( sin ⁡ 3 2 0 ∘ − cos ⁡ 3 5 0 ∘ ) 0 = 3(\sin20^{\circ}-\cos50^{\circ})-4(\sin^{3}20^{\circ}-\cos^{3}50^{\circ})

So, sin ⁡ 3 2 0 ∘ − cos ⁡ 3 5 0 ∘ sin ⁡ 2 0 ∘ − cos ⁡ 5 0 ∘ = 3 4 \frac{\sin^{3}20^{\circ}-\cos^{3}50^{\circ}}{\sin20^{\circ}-\cos50^ {\circ}}=\frac{3}{4}

which is 0.75 0.75

2 Helpful 0 Interesting 1 Brilliant 0 Confused

sin20=sin(50-30)=sin50cos30-sin30cos50=sqrt3/2sin50-1/2cos50=>. 1.(sqrt3/2sin50-1/2cos50)^2=3/4sin^2(50)+1/4cos^2(50)-sqrt3/2cos50sin50 si 2. sin20 cos50=(sqrt3/2sin50-1/2cos50) cos50 =>1.+2.=3/4

Radu Ion - 5 years, 10 months ago

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