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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > sin4x+Cos4x?
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sin4x+Cos4x?

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Dr S, please check your expansion of Cos 2x. You have expanded cos 2x as cos^2 2x - sin^2 2x which is wrong. Mistake could be by oversight. Sorry for pointing out.
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In trigonometry and in more parts of mathematics, you can not ask how much this will be as you can get and no of solutions for the same expression. So you may ask how to get the required once if you wont get that one. Please try to understand. Here we can see the above answers as well. they all are right...
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In trigonometry and in more parts of mathematics, you can not ask how much this will be as you can get and no of solutions for the same expression. So you may ask how to get the required once if you wont get that one. Please try to understand. Here we can see the above answers as well. they all are right in their own way. So what answer you need is important that too in mathematics. Thanks. read less
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K Bala, the question seems to be sin 4x + cos 4x but you have solved for Sin^4 x + cos^4 x.
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We Know that, Sin^2(x) + Cos^2(x) = 1 Squaring on both sides Sin^4(x) + Cos^4(x) + 2Sin^2(x)Cos^2(x) = 1 {Using (a + b)^2 = a^2 + b^2 + 2ab } Sin^4(x) + Cos^4(x) = 1 - 2Sin^2(x)Cos^2(x)
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sin(4x) = 2 sin(2x) cos(2x) 2 (2 sin(x) cos(x)) (cos²(x) - sin²(x)) 4 sin(x) cos(x) (cos²(x) - sin²(x)) 4 sin(x) cos³(x) - 4 sin³(x) cos(x) -------eq 1 cos(4x) 2cos^2(2x) - 1 2^2 - 1 8cos^4(x) - 8cos^2(x) + 1 -------eq 2 Add both eq1 and eq2 solve the equation in such way you want a...
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sin(4x) = 2 sin(2x) cos(2x) 2 (2 sin(x) cos(x)) (cos²(x) - sin²(x)) 4 sin(x) cos(x) (cos²(x) - sin²(x)) 4 sin(x) cos³(x) - 4 sin³(x) cos(x) -------eq 1 cos(4x) 2cos^2(2x) - 1 2[2cos^2(x) - 1]^2 - 1 8cos^4(x) - 8cos^2(x) + 1 -------eq 2 Add both eq1 and eq2 solve the equation in such way you want answer. read less
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1. Integration by parts 2. u-substitution 3. Reverse chain rule 4. Partial fraction expansion 5. Integration using trigonometric identities 6. Trigonometric substitution
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Sin(4x) + cos(4x) = Sin(2*2x) + cos(2*2x) = 2sin(2x)cos(2x) + ((2cos(2x))^2)-1 = 2(2sin(x)cos(x))((2cosx)^2-1) + (((2*2*(cosx^2)-1)^2)-1)
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sinx)^4+(cosx)^4 = 1 The soltion is obvious but a procedure is as below: s^4+(1-s^2)^2 = 1, where s = sinx. Also (sinx)^2+(cosx)^2 = 1 is a trigonometric identity for all x. So, (cosx)^4 = ^2 = (1-s^2)^2. Therefore, s^4+1-2s^2+s^4=1 2s^4-2s^2=0 s^4-s^2 = 0 s^2(s^2-1)=0 s^2 = 0 or s^2 = 1 s=0...
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sinx)^4+(cosx)^4 = 1 The soltion is obvious but a procedure is as below: s^4+(1-s^2)^2 = 1, where s = sinx. Also (sinx)^2+(cosx)^2 = 1 is a trigonometric identity for all x. So, (cosx)^4 = [(cosx)^2]^2 = (1-s^2)^2. Therefore, s^4+1-2s^2+s^4=1 2s^4-2s^2=0 s^4-s^2 = 0 s^2(s^2-1)=0 s^2 = 0 or s^2 = 1 s=0 or s=1 or s=-1 read less
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