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Geometry Level 2

If sin ⁡ x + cos ⁡ x = 1 2 \sin x + \cos x = \dfrac{1}{2} , find: sin ⁡ 3 x + cos ⁡ 3 x \sin^3 x +\cos^3 x

7/2 11/16 7/8 0.676

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Christian Daang by Christian Daang , 20, Philippines

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3 solutions

Sanjeet Raria
Oct 24, 2014

( sin ⁡ x + cos ⁡ x ) = 1 2 (\sin x+\cos x)=\frac{1}{2}

Squaring & solving further gives us, sin ⁡ x • cos ⁡ x = − 3 8 \sin x•\cos x=\frac{-3}{8} Using a famous algebraic 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 can write, ( sin ⁡ 3 x + cos ⁡ 3 x ) = ( sin ⁡ x + cos ⁡ x ) ( sin ⁡ 2 x + cos ⁡ 2 x − sin ⁡ x • cos ⁡ x (\sin^3 x +\cos^3 x)=(\sin x+\cos x)(\sin^2 x+\cos^2 x-\sin x•\cos x ⇒ ( sin ⁡ 3 x + cos ⁡ 3 x ) = 1 2 ( 1 + 3 8 ) \Rightarrow (\sin^3 x +\cos^3 x)=\frac{1}{2}(1+\frac{3}{8}) ⇒ ( sin ⁡ 3 x + cos ⁡ 3 x ) = 11 16 \Rightarrow (\Large \sin^3 x +\cos^3 x)=\boxed{\frac{11}{16}}

12 Helpful 0 Interesting 0 Brilliant 0 Confused
Christian Daang
Oct 24, 2014

Given: sin x + cos x = 1/2

Then,

(sin^2)x + 2(sin x)(cos x) + (cos^2)x = 1/4

1 + 2(sin x)(cos x) = 1/4

2(sin x)(cos x) = -3/4

(sin x)(cos x) = -3/8

And..

(sin^3)x + (cos^3)x = (sin x + cos x)[(sin^2)x - (sin x)(cos x) + (cos^2)x]

= (sin x + cos x)(1 - (sin x)(cos x))

= (1/2)(1- -3/8)

= (1/2)(11/8)

= 11/16

Final Answer: 11/16

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Sin(x)+ cos(x)= ½ 

                                   By squaring both sides, we get,

[Sin(x)+ cos(x) ]2 = (½)2

Sin2(x) + 2. Sin(x). cos(x) + cos2(x) = ¼

1 + 2. Sin(x). cos(x) = ¼ 

  2. Sin(x). cos(x) = ¼ -1

  2. Sin(x). cos(x) =  -(¾)

                                    By using      sin(2x)= 2. Sin(x).cos(x),   we get,

Sin(2x) = -(¾)

2x = sin-1 [-(¾)]

2x = -48.59

X = -24.29

By using this,value of ‘x’ we can find,

Sin3(x) + cos3(x) = sin3(-24.29) + cos3(-24.29) 

                        =  -0.0696 + 0.7571

                       = 0.6875

Sin3(x) + cos3(x) = 11/16

Pranav Deshpande - 6 years, 7 months ago
Abdul Basit
Nov 8, 2014

sinx+cosx=1/2 sin²x+cos²x+2sinxcosx=1/4 By Pythagorean identity, sin²x+cos²x=1 2sinxcosx=-3/4 sinxcosx=-3/8

Factoring sin³x+cos³x= (sinx+cosx)(sin²x-sinxcosx+cos²x) = (1/2)(1+3/8) = (1/2)(11/8) = 11/16

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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