Question

simplify.

1. $\cos 2x\cos -3x - \sin 2x\sin -3x$
2. $\sin -2x\cos 5x + \cos -2x\sin 5x$
3. $\cos 3x\cos -2x - \sin 3x\sin -2x$
4. $\dfrac{\tan 6x - \tan 5x}{1 + \tan 6x\tan 5x}$

Explanation:

Problem 7

Step1: Recall cosine addition formula

The formula for the cosine of a sum of two angles is $\cos(A + B) = \cos A\cos B - \sin A\sin B$. Also, we know that $\cos(-\theta)=\cos\theta$ and $\sin(-\theta)=-\sin\theta$. Let $A = 2x$ and $B=- 3x$.
The given expression is $\cos 2x\cos(-3x)-\sin 2x\sin(-3x)$. Substituting $\cos(-3x)=\cos 3x$ and $\sin(-3x)=-\sin 3x$ into the expression, we get $\cos 2x\cos 3x-\sin 2x(-\sin 3x)=\cos 2x\cos 3x+\sin 2x\sin 3x$. Wait, no, wait. Wait the original formula is $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Let's re - check. If we have $\cos A\cos B-\sin A\sin B$, and here $A = 2x$, $B=-3x$, then $\cos(2x+(-3x))=\cos(2x - 3x)=\cos(-x)$. And since $\cos(-x)=\cos x$, also we can use the formula directly. Let $A = 2x$ and $B = 3x$ (because $\cos(-3x)=\cos 3x$ and $\sin(-3x)=-\sin 3x$, so $-\sin 2x\sin(-3x)=-\sin 2x(-\sin 3x)=\sin 2x\sin 3x$? Wait no, let's do it step by step.

The given expression: $\cos(2x)\cos(-3x)-\sin(2x)\sin(-3x)$

We know that $\cos(-\alpha)=\cos\alpha$ and $\sin(-\alpha)=-\sin\alpha$. So $\cos(-3x)=\cos(3x)$ and $\sin(-3x)=-\sin(3x)$

Substitute these into the expression:

$\cos(2x)\cos(3x)-\sin(2x)(-\sin(3x))=\cos(2x)\cos(3x)+\sin(2x)\sin(3x)$

Wait, but the cosine addition formula is $\cos(A - B)=\cos A\cos B+\sin A\sin B$. Let $A = 2x$ and $B = 3x$, then $\cos(2x-3x)=\cos(-x)=\cos x$. Alternatively, using $\cos(A + B)=\cos A\cos B-\sin A\sin B$, if we let $A = 2x$ and $B=-3x$, then $\cos(2x+(-3x))=\cos(2x)\cos(-3x)-\sin(2x)\sin(-3x)$, which is exactly the given expression. So $\cos(2x-3x)=\cos(-x)=\cos x$ (since $\cos(-\theta)=\cos\theta$)

Step2: Simplify the angle

$2x-3x=-x$, so $\cos(2x - 3x)=\cos(-x)=\cos x$

Step1: Recall sine addition formula

The formula for the sine of a sum of two angles is $\sin(A + B)=\sin A\cos B+\cos A\sin B$. We know that $\sin(-\theta)=-\sin\theta$ and $\cos(-\theta)=\cos\theta$. Let $A=-2x$ and $B = 5x$.

The given expression is $\sin(-2x)\cos(5x)+\cos(-2x)\sin(5x)$

Since $\sin(-2x)=-\sin(2x)$ and $\cos(-2x)=\cos(2x)$, substitute these into the expression:

$(-\sin(2x))\cos(5x)+\cos(2x)\sin(5x)=\cos(2x)\sin(5x)-\sin(2x)\cos(5x)$

Wait, no, the sine addition formula is $\sin(A + B)=\sin A\cos B+\cos A\sin B$. Let's take $A=-2x$ and $B = 5x$, then $\sin(-2x + 5x)=\sin(3x)$. Let's check:

$\sin(A + B)=\sin A\cos B+\cos A\sin B$, where $A=-2x$ and $B = 5x$

$\sin(-2x+5x)=\sin(-2x)\cos(5x)+\cos(-2x)\sin(5x)$

Which is exactly the given expression. So $\sin(-2x + 5x)=\sin(3x)$

Step2: Simplify the angle

$-2x + 5x=3x$, so the expression simplifies to $\sin(3x)$

Step1: Recall cosine addition formula

The formula for the cosine of a sum of two angles is $\cos(A + B)=\cos A\cos B-\sin A\sin B$. We know that $\cos(-\alpha)=\cos\alpha$ and $\sin(-\alpha)=-\sin\alpha$. Let $A = 3x$ and $B=-2x$

The given expression is $\cos(3x)\cos(-2x)-\sin(3x)\sin(-2x)$

Since $\cos(-2x)=\cos(2x)$ and $\sin(-2x)=-\sin(2x)$, substitute these into the expression:

$\cos(3x)\cos(2x)-\sin(3x)(-\sin(2x))=\cos(3x)\cos(2x)+\sin(3x)\sin(2x)$

But the formula for $\cos(A - B)=\cos A\cos B+\sin A\sin B$. Let $A = 3x$ and $B = 2x$, then $\cos(3x-2x)=\cos(x)$

Alternatively, using the formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$ with $A = 3x$ and $B=-2x$, we have $\cos(3x+(-2x))=\cos(3x - 2x)=\cos(x)$

Step2: Simplify the angle

$3x-2x=x$, so the expression simplifies to $\cos(x)$

Answer:

$\cos x$

Problem 8