sin(n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos ( ( n + 2 ) x − ( n − 1 ) x ) { ∵ cos ( A − B ) = sin A sin B + cos A cos B } ⇒ = cos ( ( n + 2 − n − 1 ) x ) Soif X n is the number of individuals alive in generation n, then X n+1 is the sum of X n -many independent, identically distributed random variables. Let's assume that X 0 = 1, p (0) > 0, and = k p (k) = E (X 1) 1. (a) If = 1 and 2 < , then there exist constants 0 < c 1 < c 2 < such that. c 1 /n < P ( X n 0 ) < c 2 /n. S.}=sin (n+1) x sin (n+2) x+cos (n+1) x (n+2) x ) ( begin{aligned} &=text { Put }(n+1) x=A text { and }(n+2) x=mathbf{B} therefore text { L. } mathbf{H . S .} &=sin A sin B+cos A cos B &=cos A cos B+sin A sin B=cos (A-B) &=cos [(n+1) x-(n+2) x]=cos (n x+x-n x-2 x) &=cos (-x)=cos x=R . If[{Sin[(n + 1)x] + Sinx}/x] for lim x→0 = (1/2) then value of n is: (a) - 2.5 (b) - 0.5 (c) - 1.5 (d) - 1 Multiplyboth sides by n n. 1+sin(x) n n = kn 1 + sin ( x) n n = k n. Simplify the left side. Tap for more steps sin(x)+1 = kn sin ( x) + 1 = k n. Subtract 1 1 from both sides of the equation. sin(x) = kn−1 sin ( x) = k n - 1. Take the inverse sine of both sides of the equation to extract x x from inside the sine. Inthe particular case of your question, we have the simple algebraic identity $$(n+1)x=nx+x.$$ When applying the sine function to this quantity, we need no further parentheses on the left-hand side; but parentheses must be introduced on the right-hand side because otherwise it would read as $$\sin nx+x,$$ which is the sum of $\sin nx$ and $x$. Wy9Bb9. $\begingroup$ Question Prove that $\sinnx \cosn+1x-\sinn-1x\cosnx = \sinx \cos2nx$ for $n \in \mathbb{R}$. My attempts I initially began messing around with the product to sum identities, but I couldn't find any way to actually use them. I also tried compound angles to expand the expression, but it became too difficult to work with. Any help or guidance would be greatly appreciated asked Jun 15, 2020 at 1531 $\endgroup$ 2 $\begingroup$The left-hand side is$$\begin{align}&\sin nx\cos nx\cos x-\sin nx\sin x-\sin nx\cos x-\cos nx\sin x\cos nx\\&=\cos^2nx-\sin^2nx\sin x\\&=\cos 2nx\sin x.\end{align}$$ answered Jun 15, 2020 at 1537 gold badges74 silver badges135 bronze badges $\endgroup$ $\begingroup$Use $\sina\cosb=\frac{1}{2}\sina-b+\sina+b$ $$ \sinnx \cosn+1x-\sinn-1x\cosnx $$ $$ =\frac{1}{2}\left\sin-x+\sin2n+1x-\sin-x-\sin2n-1x \right $$ $$ =\frac{1}{2}\left\sin2n+1x-\sin2n-1x \right $$Now use $\sina+b=\sina\cosb+\sinb\cosa$ $$ =\frac{1}{2}\left\sin2nx\cosx +\sinx\cos2nx-\sin2nx\cos-x-\sin-x\cos2nx \right $$Now use the parity of sine and cosine and you're done. $$ =\frac{1}{2}\left\sin2nx\cosx +\sinx\cos2nx-\sin2nx\cosx+\sinx\cos2nx \right $$ $$ =\sinx\cos2nx $$ answered Jun 15, 2020 at 1536 IntegrandIntegrand8,15415 gold badges41 silver badges69 bronze badges $\endgroup$ $\begingroup$ $$ \begin{align} \sinnx\cosn+1x &=\frac{\sinnx+n+1x+\sinnx-n+1x}2\tag1\\ &=\frac{\sin2n+1x-\sinx}2\tag2\\ \sinn-1x\cosnx &=\frac{\sin2n-1x-\sinx}2\tag3 \end{align} $$ Explanation $1$ identity $\sina\cosb=\frac{\sina+b+\sina-b}2$ $2$ simplify $3$ apply $2$ for $n-1$ Therefore, $$ \begin{align} \sinnx\cosn+1x-\sinn-1x\cosnx &=\frac{\sin2n+1x-\sin2n-1x}2\tag4\\ &=\sinx\cos2nx\tag5 \end{align} $$ Explanation $4$ subtract $3$ from $2$ $5$ identity $\sina-\sinb=2\sin\left\frac{a-b}2\right\cos\left\frac{a+b}2\right$ answered Jun 15, 2020 at 1822 robjohn♦robjohn337k35 gold badges446 silver badges832 bronze badges $\endgroup$ You must log in to answer this question. Not the answer you're looking for? Browse other questions tagged . Calculus Examples Popular Problems Calculus Solve for x k=1+sinx/n Step 1Rewrite the equation as .Step 2Multiply both sides by .Step 3Simplify the left for more steps...Step .Tap for more steps...Step the common factor of .Tap for more steps...Step the common the and .Step 4Subtract from both sides of the 5Take the inverse sine of both sides of the equation to extract from inside the sine. Trigonometry Examples Popular Problems Trigonometry Simplify 1+sinx1-sinx Step 1Apply the distributive 2Multiply by .Step 3Rewrite using the commutative property of 4Multiply .Tap for more steps...Step to the power of .Step to the power of .Step the power rule to combine and .

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