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sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=co...

`sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx`

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To solve the equation \( \sin((n+1)x) \sin((n+2)x) + \cos((n+1)x) \cos((n+2)x) = \cos(x) \), we can use the cosine addition formula. Let's break it down step by step. ### Step 1: Use the Cosine Addition Formula We know from trigonometric identities that: \[ \cos(a - b) = \cos a \cos b + \sin a \sin b \] In our case, we can set \( a = (n+1)x \) and \( b = (n+2)x \). Thus, we can rewrite the left-hand side of the equation as: ...
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Let I_(m","n)= int sin^(n)x cos^(m)x dx . Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) " " (ii) I_(n+2","m) (iii) I_(n","m-2) " " (iv) I_(n","m+2) (v) I_(n-2","m+2)" " I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2) , then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos x in m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1 . Now, choose the exponent of sin x for m - 1 of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x . Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation between I_(4","2) and I_(2","2) is

Let I_(m","n)= int sin^(n)x cos^(m)x dx . Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) " " (ii) I_(n+2","m) (iii) I_(n","m-2) " " (iv) I_(n","m+2) (v) I_(n-2","m+2)" " I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2) , then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos x in m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1 . Now, choose the exponent of sin x for m - 1 of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x . Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation between I_(4","2) and I_(6","2) is

Let I_(m","n)= int sin^(n)x cos^(m)x dx . Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) " " (ii) I_(n+2","m) (iii) I_(n","m-2) " " (iv) I_(n","m+2) (v) I_(n-2","m+2)" " I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2) , then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos x in m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1 . Now, choose the exponent of sin x for m - 1 of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x . Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation I_(4","2) and I_(4","4) is

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