Plzz Solve fast (Trigonometry )sum::
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Given 2(sin^6 theta + cos^6 theta) - 3(sin^4 theta + cos^4 theta) + 1 ----------- (1)
Let's divide the equation into 2 parts in order to avoid confusion.
1st Part :
2(sin^6 theta + cos^6 theta) can be written as
2((sin^2 theta)^3 + (cos^2 theta)^3).
We know that (a^3 + b^3) = (a + b)(a^2 - ab + b^2).
2(sin ^2 theta + cos^2 theta)(sin^4 theta - sin^2 thetacos^2 theta + cos^4 theta)
= 2(1)(sin^4 theta - sin^2 theta cos^2 theta + cos^4 theta)
= 2(sin^4 theta - sin^2 theta cos^2 theta + cos^4 theta)
= 2sin^4 theta - 2sin^2 theta cos^2 theta + 2 cos^4 theta --------- (2)
2nd Part :
-3(sin^4 theta + cos^4 theta) + 1 ---------- (3)
Substitute (2) & (3) in (1), we get
= 2sin^4 theta - 2 sin^2 theta cos^2 theta + 2 cos^4 theta - 3 sin^4 theta - 3 cos^4 theta + 1
= -sin^4 theta - cos^4 theta - 2 sin^2 theta cos^2 theta + 1
= -(sin^4 theta + cos^4 theta + 2 sin^2 theta cos^2 theta) + 1
= -(sin^2 theta + cos^2 theta)^2 + 1
= -1 + 1
= 0.
Hence proved.
Hope this helps!
Let's divide the equation into 2 parts in order to avoid confusion.
1st Part :
2(sin^6 theta + cos^6 theta) can be written as
2((sin^2 theta)^3 + (cos^2 theta)^3).
We know that (a^3 + b^3) = (a + b)(a^2 - ab + b^2).
2(sin ^2 theta + cos^2 theta)(sin^4 theta - sin^2 thetacos^2 theta + cos^4 theta)
= 2(1)(sin^4 theta - sin^2 theta cos^2 theta + cos^4 theta)
= 2(sin^4 theta - sin^2 theta cos^2 theta + cos^4 theta)
= 2sin^4 theta - 2sin^2 theta cos^2 theta + 2 cos^4 theta --------- (2)
2nd Part :
-3(sin^4 theta + cos^4 theta) + 1 ---------- (3)
Substitute (2) & (3) in (1), we get
= 2sin^4 theta - 2 sin^2 theta cos^2 theta + 2 cos^4 theta - 3 sin^4 theta - 3 cos^4 theta + 1
= -sin^4 theta - cos^4 theta - 2 sin^2 theta cos^2 theta + 1
= -(sin^4 theta + cos^4 theta + 2 sin^2 theta cos^2 theta) + 1
= -(sin^2 theta + cos^2 theta)^2 + 1
= -1 + 1
= 0.
Hence proved.
Hope this helps!
Anonymous:
Thank you so much
My Pleasure
Thnx u again u helped a lot tomorrow is my exam
Good Luck for your exam
Thanks Mahek For The Brainliest
It was my pleasure u helped me !!!
:-)
Answered by
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2(sin^6theta +cos^6theta) -3(sin^4theta+ cos^4theta) +1=0
2sin^6theta+2cos^6theta-3sin^4theta-3cos^4theta+1=0
(-sin^2theta-cos^2theta)+1=0
-(sin^2theta+cos^2theta) +1=0
(-(sin^2theta+1/sin^2theta)) +1=0
-1+1=0
0=0
hence proved....
2sin^6theta+2cos^6theta-3sin^4theta-3cos^4theta+1=0
(-sin^2theta-cos^2theta)+1=0
-(sin^2theta+cos^2theta) +1=0
(-(sin^2theta+1/sin^2theta)) +1=0
-1+1=0
0=0
hence proved....
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