Question

How do you simplify $\sin (70)\cos (40)-\cos (70)\sin (40)$?

Answer 1

Here, in this problem, we need to know about the trigonometry and its property that is $\sin (A-B)=\sin (A)\cos (B)-\cos (A)\sin (B)$ in this property we have to use for this particular problem. By substituting the value of A and B we get the required answer.

Complete answer:
Before we solve this problem, first we need to know about trigonometry and its identity.
So, first of all we need to know about the trigonometry that is
Trigonometry is the branch of mathematics which is basically concerned with specific functions of angles, their applications and their calculations. The trigonometric functions are very important for studying triangles, light, sound or waves.
Now we will see about trigonometry identity that is sum and difference formula of different trigonometric functions are as follows:

1. $\sin (A+B)=\sin (A)\cos (B)+\cos (A)\sin (B)$
2. $\sin (A-B)=\sin (A)\cos (B)-\cos (A)\sin (B)$
3. $\cos (A+B)=\cos (A)\cos (B)-\sin (A)\sin (B)$
4. $\cos (A-B)=\cos (A)\cos (B)+\sin (A)\sin (B)$
   Out of this four trigonometry we will likely to discuss about the second one which is given by $\Rightarrow \sin (A-B)=\sin (A)\cos (B)-\cos (A)\sin (B)--(1)$
   Because if you observe the question which is given by
   $\Rightarrow \sin (70)\cos (40)-\cos (70)\sin (40)---(2)$
   That is above problem matches with second one of the trigonometry identities
   By comparing equation (2) with equation (1) we get the value of A and B that is $A=70$ and $B=40$
   By substituting this value on Formula that is equation (1) we get:
   $\Rightarrow \sin (70)\cos (40)-\cos (70)\sin (40)=\sin (70-40)$
   By simplifying this we get:
   $\Rightarrow \sin (70)\cos (40)-\cos (70)\sin (40)=\sin (30)$
   As we know that $\sin (30)=\dfrac{1}{2}$ and substitute in the above equation we get:
   $\Rightarrow \sin (70)\cos (40)-\cos (70)\sin (40)=\dfrac{1}{2}$
   Hence, the value of $\sin (70)\cos (40)-\cos (70)\sin (40)$ is $\dfrac{1}{2}$.

Note:
In this problem, we have to remember all the trigonometry identity other than one which we used in this problem, and also use the trigonometry values of each trigonometric functions for example $\sin (30)=\dfrac{1}{2}$, $\sin (90)=1$ similar type should remember for cosine sec and tan. Always remember that the maximum value of the trigonometric function is 1. You may also require the trigonometry ratio of the right-angled triangle in similar types of problems. For example, the sine of an angle of a right-angled triangle is the ratio of its perpendicular and hypotenuse. This can be written as $\sin (x)=\dfrac{P}{H}$ . Similarly, the cosine of an angle of a right-angled triangle is the ratio of its base and hypotenuse. This can be written as $\cos (x)=\dfrac{B}{H}$. The secant of an angle of a right-angled triangle is the ratio of its hypotenuse and base. This can be written as $\sec (x)=\dfrac{H}{B}$. The cosecant of an angle of a right-angled triangle is the ratio of its hypotenuse and perpendicular. This can be written as $\csc (x)=\dfrac{H}{P}$. So, this type of formula you have to remember because the Trigonometry ratio formula is based on a question asked in the exam.