Trigonometric Identities

Step 1: Understanding the inverse sine and cosine functions.
Let $y%20%3D%20%5Csin%5E%7B-1%7D(x)$ . By the definition of inverse sine: $%5Csin(y)%20%3D%20x%20%5C%2C%5C%2C%5C%2C%20%5Ctext%7Bwhere%7D%20%5C%2C%5C%2C%5C%2C%20y%20%5Cin%20%5Cleft%5B%20-%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5Cright%5D$ Similarly, let $z%20%3D%20%5Ccos%5E%7B-1%7D(x)$ . By the definition of inverse cosine: $%5Ccos(z)%20%3D%20x%20%5C%2C%5C%2C%5C%2C%20%5Ctext%7Bwhere%7D%20%5C%2C%5C%2C%5C%2C%20z%20%5Cin%20%5Cleft%5B%200%2C%20%5Cpi%20%5Cright%5D$

Step 2: Proof of the identity.
We know that for all $x$ in the interval $%5B-1%2C%201%5D$ , $%5Csin%5E%7B-1%7D(x)%20%2B%20%5Ccos%5E%7B-1%7D(x)%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D$ . To prove this, observe the following: - The sine and cosine functions are complementary, meaning $%5Csin%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D%20-%20x%5Cright)%20%3D%20%5Ccos(x)$ . - Since $%5Csin%5E%7B-1%7D(x)%20%3D%20y$ and $%5Ccos%5E%7B-1%7D(x)%20%3D%20z$ , we know that $%5Csin(y)%20%3D%20x$ and $%5Ccos(z)%20%3D%20x$ . - By definition, $%5Csin%5E%7B-1%7D(x)%20%2B%20%5Ccos%5E%7B-1%7D(x)%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D$ because $y%20%2B%20z%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D$ .

Step 3: Conclusion.
Thus, we have proved that $%5Csin%5E%7B-1%7D(x)%20%2B%20%5Ccos%5E%7B-1%7D(x)%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D$ for $x%20%5Cin%20%5B-1%2C%201%5D$ .

Step 1: Applying the addition formula for inverse tangents.
We can use the formula for the sum of inverse tangents: $%5Ctan%5E%7B-1%7D%20a%20%2B%20%5Ctan%5E%7B-1%7D%20b%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft(%5Cfrac%7Ba%20%2B%20b%7D%7B1%20-%20ab%7D%5Cright)%2C$ if $ab%20%3C%201$ . Substituting $a%20%3D%20%5Cfrac%7B1%7D%7B2%7D$ and $b%20%3D%20%5Cfrac%7B1%7D%7B3%7D$ : $%5Ctan%5E%7B-1%7D%20%5Cfrac%7B1%7D%7B2%7D%20%2B%20%5Ctan%5E%7B-1%7D%20%5Cfrac%7B1%7D%7B3%7D%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft(%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%20%2B%20%5Cfrac%7B1%7D%7B3%7D%7D%7B1%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%7D%5Cright).$

Step 2: Simplifying the expression.
The numerator is: $%5Cfrac%7B1%7D%7B2%7D%20%2B%20%5Cfrac%7B1%7D%7B3%7D%20%3D%20%5Cfrac%7B3%20%2B%202%7D%7B6%7D%20%3D%20%5Cfrac%7B5%7D%7B6%7D.$ The denominator is: $1%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%20%3D%201%20-%20%5Cfrac%7B1%7D%7B6%7D%20%3D%20%5Cfrac%7B5%7D%7B6%7D.$ Thus: $%5Ctan%5E%7B-1%7D%20%5Cfrac%7B1%7D%7B2%7D%20%2B%20%5Ctan%5E%7B-1%7D%20%5Cfrac%7B1%7D%7B3%7D%20%3D%20%5Ctan%5E%7B-1%7D%20%5Cleft(%5Cfrac%7B%5Cfrac%7B5%7D%7B6%7D%7D%7B%5Cfrac%7B5%7D%7B6%7D%7D%5Cright)%20%3D%20%5Ctan%5E%7B-1%7D%201.$ Since $%5Ctan%5E%7B-1%7D%201%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D$ , the equation is false.

Step 3: Conclusion.
Thus, $%5Ctan%5E%7B-1%7D%20%5Cfrac%7B1%7D%7B2%7D%20%2B%20%5Ctan%5E%7B-1%7D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cneq%20%5Ctan%5E%7B-1%7D%20%5Cfrac%7B1%7D%7B5%7D$ , making the statement false.

High-Yield Trend

Chapter Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Trigonometric Identities

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Chapter Questions
3 MCQs