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# How do you use the sum and difference formula to simplify $\sin 40\cos 10+\cos 40\sin 10$ ?

Last updated date: 20th Jun 2024
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Hint:
We apply the theorem of the formula of sum and difference of angles for sin ratio. The formula is $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ . We put the values of $A=40; B=10$ in the equation to find the sum of the angles and the solution to the problem.

We know the formula of sum and difference of angles as
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ .
For our given problem we use the value of the angles as $A=40; B=10$ .
Placing the values, we get
$\sin 40\cos 10+\cos 40\sin 10=\sin \left( 40+10 \right)=\sin 50$ .
We also know that $\sin \alpha =\cos \left( 90-\alpha \right)$ . We put the value of $\alpha$ as 50.
So, we can also write that $\sin 50=\cos \left( 90-50 \right)=\cos 40$ .
Therefore, the value of $\sin 40\cos 10+\cos 40\sin 10$ is $\sin 50$ .

Note:
We can also covert the sin ratios in cos ratios and make the equation in such a way that we can apply another formula of sum and difference of ratio cos to find the value of the problem. The formula is $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ .
We change $\sin 40\cos 10+\cos 40\sin 10=\cos 50\cos 10+\sin 50\sin 10$ , using the formula of $\sin \alpha =\cos \left( 90-\alpha \right)$ .
We put the values of $A=50;B=10$
So, $\cos 50\cos 10+\sin 50\sin 10=\cos \left( 50-10 \right)=\cos 40$ . The solution is same.