A geometry problem by Mehul Chaturvedi

Geometry Level 3

F i n d c o s 2 A + c o s 2 ( A + 120 ) + c o s 2 ( A − 120 ) Find\\ { cos }^{ 2 }A+{ cos }^{ 2 }(A+120)+{ cos }^{ 2 }(A-120)


The answer is 1.5.

Mehul Chaturvedi by Mehul Chaturvedi , 22, India

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3 solutions

Jaiveer Shekhawat
Oct 10, 2014

c o s 2 A cos^{2}A + c o s 2 ( A + 120 ) cos^{2}(A+120) + c o s 2 ( A − 120 ) cos^{2}(A-120)

= c o s 2 A cos^{2}A + ( c o s A c o s 120 − s i n A s i n 120 ) 2 (cosAcos120-sinAsin120)^{2} + ( c o s A c o s 120 + s i n A s i n 120 ) 2 (cosAcos120+sinAsin120)^{2}

= c o s 2 A cos^{2}A + ( c o s A c o s ( 90 + 30 ) − s i n A s i n ( 180 − 60 ) ) 2 (cosAcos(90+30)-sinAsin(180-60))^{2} + ( c o s A c o s ( 90 + 30 ) + s i n A s i n ( 180 − 60 ) ) 2 (cosAcos(90+30)+sinAsin(180-60))^{2}

= c o s 2 A cos^{2}A + ( − c o s A s i n 30 − s i n A s i n 60 ) 2 (-cosAsin30 - sinAsin60)^{2} + ( − c o s A s i n 30 + s i n A s i n 60 ) 2 (-cosAsin30 + sinAsin60)^{2}

= c o s 2 A cos^{2}A + ( c o s A s i n 30 ) 2 (cosAsin30)^{2} -2(-cosAsin30)(sinAsin60) + ( s i n A s i n 60 ) 2 sinAsin60)^{2} + ( c o s A s i n 30 ) 2 (cosAsin30)^{2} +2(-cosAsin30)(sinAsin60) + ( s i n A s i n 60 ) 2 sinAsin60)^{2}

= c o s 2 A cos^{2}A + ( c o s A s i n 30 ) 2 (cosAsin30)^{2} + s i n A s i n 60 ) 2 sinAsin60)^{2} + ( c o s A s i n 30 ) 2 (cosAsin30)^{2} + ( s i n A s i n 60 ) 2 sinAsin60)^{2}

= c o s 2 A cos^{2}A + 2. c o s 2 A cos^{2}A . 1 4 \frac{1}{4} + 2. s i n 2 A sin^{2}A . 3 4 \frac{3}{4}

= c o s 2 A cos^{2}A + c o s 2 A cos^{2}A . 1 2 \frac{1}{2} + ( 1 − c o s 2 A 1-cos^{2}A ). 3 2 \frac{3}{2}

=2. 1 2 \frac{1}{2} c o s 2 A cos^{2}A + c o s 2 A cos^{2}A . 1 2 \frac{1}{2} + 3 2 \frac{3}{2} + − c o s 2 A -cos^{2}A . 3 2 \frac{3}{2}

= c o s 2 A cos^{2}A . 3 2 \frac{3}{2} + − c o s 2 A -cos^{2}A . 3 2 \frac{3}{2} + 3 2 \frac{3}{2}

= 3 2 \frac{3}{2}

= 1.5 \boxed{1.5}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I have explained each step one by one for reader's comfort!

jaiveer shekhawat - 6 years, 8 months ago

easy solution----= check for A=0

1 Helpful 0 Interesting 0 Brilliant 0 Confused
James Villanueva
Oct 13, 2014

c o s 2 A + c o s 2 ( A + 120 ) + c o s 2 ( A − 120 ) cos^2A+cos^2(A+120)+cos^2(A-120)

= c o s 2 A + [ c o s ( A + 120 ) + c o s ( A − 120 ) ] 2 − 2 c o s ( A + 120 ) c o s ( A − 120 ) =cos^2A+[cos(A+120)+cos(A-120)]^2-2cos(A+120)cos(A-120)

Now using the sum to product identity in the second term and the product to sum identity in the third term, we get:

= c o s 2 A + ( 2 c o s [ ( A + 120 ) + ( A − 120 ) 2 ] c o s [ ( A + 120 ) − ( A − 120 ) 2 ] ) 2 − 2 ( c o s [ ( A + 120 ) + ( A − 120 ) ] + c o s [ ( A + 120 ) − ( A − 120 ) ] 2 ) =cos^2A+\left(2cos \left[\frac{(A+120)+(A-120)}{2}\right] cos \left[\frac{(A+120)-(A-120)}{2}\right]\right)^2-2 \left(\frac{cos[(A+120)+(A-120)]+cos[(A+120)-(A-120)]}{2}\right)

Simplifying the angles of each cosine:

= c o s 2 A + ( 2 c o s A c o s 120 ) 2 − [ c o s ( 2 A ) + c o s 240 ] =cos^2A+\left(2cosAcos120 \right)^2-[cos(2A)+cos240]

Solving the square in the second term and distributing the minus over the third term:

= c o s 2 A + c o s 2 A − c o s ( 2 A ) − c o s 240 =cos^2A+cos^2A-cos(2A)-cos240

Using the double angle identity for cosine:

= 2 c o s 2 A − ( c o s 2 A − s i n 2 A ) − ( − 1 2 ) =2cos^2A-(cos^2A - sin^2A)-\left (-\displaystyle \frac{1}{2} \right)

Combininig like terms:

= c o s 2 A + s i n 2 A + 1 2 =cos^2A+sin^2A+\displaystyle \frac{1}{2}

Using the pythagorean identity:

= 1 + 1 2 =1+\displaystyle \frac{1}{2}

Solving the operations:

= 3 2 =\displaystyle \frac{3}{2}

= 1.5 =\boxed{1.5}

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