Question

express the following as a function of a single angle.
cos 60° cos(-20°) - sin 60° sin(-20°)
sum & difference formulas
sine and cosine
cos(a ± b) =
cosacosb ∓ sinasinb
sin(a ± b) =
sinacosb ± cosasinb
options:
sin 80°
cos 40°
cos 80°
sin 40°

Explanation:

Step1: Identify the formula

The given expression is $\cos 60^{\circ}\cos(-20^{\circ}) - \sin 60^{\circ}\sin(-20^{\circ})$. We use the cosine sum formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Here, $A = 60^{\circ}$ and $B=- 20^{\circ}$.

Step2: Apply the formula

Substitute $A = 60^{\circ}$ and $B =-20^{\circ}$ into the formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$. So we get $\cos(60^{\circ}+(- 20^{\circ}))=\cos(60^{\circ}-20^{\circ})=\cos40^{\circ}$? Wait, no, wait: $A + B=60^{\circ}+(-20^{\circ}) = 40^{\circ}$? Wait, no, wait the formula is $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Wait, the given expression is $\cos A\cos B-\sin A\sin B$ where $A = 60^{\circ}$, $B=-20^{\circ}$. So $\cos(A + B)=\cos(60^{\circ}+(-20^{\circ}))=\cos(40^{\circ})$? Wait, no, wait let's check again. Wait, the formula is $\cos(A + B)=\cos A\cos B-\sin A\sin B$. So if we have $\cos A\cos B-\sin A\sin B$, that is $\cos(A + B)$. So $A = 60^{\circ}$, $B=-20^{\circ}$, so $A + B=60^{\circ}+(-20^{\circ})=40^{\circ}$? Wait, but wait, let's compute $60^{\circ}+(-20^{\circ}) = 40^{\circ}$, so $\cos(40^{\circ})$? Wait, no, wait the options have $\cos80^{\circ}$, $\cos40^{\circ}$, etc. Wait, maybe I made a mistake. Wait, let's re - examine the formula. Wait, the formula is $\cos(A - B)=\cos A\cos B+\sin A\sin B$, and $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Wait, the given expression is $\cos60^{\circ}\cos(-20^{\circ})-\sin60^{\circ}\sin(-20^{\circ})$. Let's recall that $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$. So $\cos(-20^{\circ})=\cos20^{\circ}$, $\sin(-20^{\circ})=-\sin20^{\circ}$. Substitute these into the expression: $\cos60^{\circ}\cos20^{\circ}-\sin60^{\circ}(-\sin20^{\circ})=\cos60^{\circ}\cos20^{\circ}+\sin60^{\circ}\sin20^{\circ}$. Now, this is the formula for $\cos(A - B)=\cos A\cos B+\sin A\sin B$, where $A = 60^{\circ}$, $B = 20^{\circ}$. So $\cos(60^{\circ}-20^{\circ})=\cos40^{\circ}$? Wait, no, wait $A - B=60^{\circ}-20^{\circ}=40^{\circ}$, so $\cos40^{\circ}$? But wait, let's check the original formula again. Wait, the original expression is $\cos60^{\circ}\cos(-20^{\circ})-\sin60^{\circ}\sin(-20^{\circ})$. Let's use the formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$ with $A = 60^{\circ}$, $B=-20^{\circ}$. Then $A + B=60^{\circ}+(-20^{\circ}) = 40^{\circ}$, so $\cos(40^{\circ})$? But wait, the options include $\cos80^{\circ}$. Wait, maybe I messed up the signs. Wait, $\sin(-20^{\circ})=-\sin20^{\circ}$, so $-\sin60^{\circ}\sin(-20^{\circ})=-\sin60^{\circ}(-\sin20^{\circ})=\sin60^{\circ}\sin20^{\circ}$. So the expression becomes $\cos60^{\circ}\cos(-20^{\circ})+\sin60^{\circ}\sin20^{\circ}$. Since $\cos(-20^{\circ})=\cos20^{\circ}$, this is $\cos60^{\circ}\cos20^{\circ}+\sin60^{\circ}\sin20^{\circ}$, which is $\cos(60^{\circ}-20^{\circ})=\cos40^{\circ}$? Wait, no, $\cos(A - B)=\cos A\cos B+\sin A\sin B$, so yes, $A = 60^{\circ}$, $B = 20^{\circ}$, so $\cos(60 - 20)=\cos40^{\circ}$. But wait, let's check the other way. Wait, maybe the formula is $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Let's take $A = 60^{\circ}$, $B = 20^{\circ}$, then $\cos(60 + 20)=\cos80^{\circ}=\cos60^{\circ}\cos20^{\circ}-\sin60^{\circ}\sin20^{\circ}$. But in our expression, we have $\cos60^{\circ}\cos(-20^{\circ})-\sin60^{\circ}\sin(-20^{\circ})$. Since $\cos(-20^{\circ})=\cos20^{\circ}$ and $\sin(-20^{\circ})=-\sin20^{\circ}$, so $-\sin60^{\circ}\sin(-20^{\circ})=-\sin60^{\circ}(-\sin20^{\circ})=\sin60^{\circ}\sin20^{\circ}$. So the expression is $\cos60^{\circ}\cos20^{\circ}+\sin60^{\circ}\sin20^{\circ}=\cos(60 - 20)=\cos40^{\circ}$? Wait, no, I think I made a mistake. Wait, let's start over. The given expression is $\cos60^{\circ}\cos(-20^{\circ})-\sin60^{\circ}\sin(-20^{\circ})$. The cosine sum formula is $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Let's let $A = 60^{\circ}$ and $B=-20^{\circ}$. Then $\cos(A + B)=\cos(60^{\circ}+(-20^{\circ}))=\cos(40^{\circ})$? But wait, $\cos(60^{\circ}+(-20^{\circ}))=\cos(40^{\circ})$, and the expression is exactly $\cos A\cos B-\sin A\sin B$ with $A = 60^{\circ}$ and $B=-20^{\circ}$. So the expression simplifies to $\cos(60^{\circ}+(-20^{\circ}))=\cos(40^{\circ})$? But wait, the options have $\cos80^{\circ}$. Wait, maybe I misread the formula. Wait, the formula for $\cos(A - B)$ is $\cos A\cos B+\sin A\sin B$, and $\cos(A + B)$ is $\cos A\cos B-\sin A\sin B$. Let's check with $A = 60^{\circ}$ and $B = 20^{\circ}$. $\cos(60 + 20)=\cos80^{\circ}=\cos60^{\circ}\cos20^{\circ}-\sin60^{\circ}\sin20^{\circ}$. Now, in our problem, we have $\cos60^{\circ}\cos(-20^{\circ})-\sin60^{\circ}\sin(-20^{\circ})$. Since $\cos(-20^{\circ})=\cos20^{\circ}$ and $\sin(-20^{\circ})=-\sin20^{\circ}$, so substituting, we get $\cos60^{\circ}\cos20^{\circ}-\sin60^{\circ}(-\sin20^{\circ})=\cos60^{\circ}\cos20^{\circ}+\sin60^{\circ}\sin20^{\circ}$, which is $\cos(60 - 20)=\cos40^{\circ}$. Wait, this is a contradiction. Wait, no, let's compute the numerical values. $\cos60^{\circ}=0.5$, $\cos(-20^{\circ})\approx0.9397$, $\sin60^{\circ}\approx0.8660$, $\sin(-20^{\circ})\approx - 0.3420$. So the expression is $0.5\times0.9397-0.8660\times(-0.3420)=0.46985 + 0.296172 = 0.766022$. Now, $\cos40^{\circ}\approx0.7660$, $\cos80^{\circ}\approx0.1736$, $\sin80^{\circ}\approx0.9848$, $\sin40^{\circ}\approx0.6428$. So the value of the expression is approximately equal to $\cos40^{\circ}$. Wait, but according to the formula, when we use $\cos(A + B)$ with $A = 60^{\circ}$ and $B=-20^{\circ}$, we get $\cos(40^{\circ})$, which matches the numerical calculation. So the expression simplifies to $\cos40^{\circ}$. Wait, but let's check the formula again. The formula is $\cos(A + B)=\cos A\cos B-\sin A\sin B$. So if $B=-20^{\circ}$, then $\cos(A + B)=\cos A\cos(-B)-\sin A\sin(-B)$. Since $\cos(-B)=\cos B$ and $\sin(-B)=-\sin B$, this becomes $\cos A\cos B+\sin A\sin B$, which is $\cos(A - B)$. Wait, so $\cos(A + (-B))=\cos(A - B)=\cos A\cos B+\sin A\sin B$. But in our case, the expression is $\cos A\cos B-\sin A\sin B$ with $B=-20^{\circ}$, so $A = 60^{\circ}$, $B=-20^{\circ}$, so $\cos(A + B)=\cos(60^{\circ}+(-20^{\circ}))=\cos40^{\circ}$, and the expression is $\cos A\cos B-\sin A\sin B$ which is the formula for $\cos(A + B)$. So that's correct. So the expression simplifies to $\cos(60^{\circ}+(-20^{\circ}))=\cos40^{\circ}$. Wait, but the numerical value matches $\cos40^{\circ}$. So the answer should be $\cos40^{\circ}$.

Answer:

$\cos 40^{\circ}$ (corresponding to the option: cos 40°)