Exercise

$\sec\left(x\right)=\frac{\sin\left(2x\right)}{\sin\left(x\right)}-\frac{\cos\left(2x\right)}{\cos\left(x\right)}$

Step-by-step Solution

1

Starting from the right-hand side (RHS) of the identity

$\frac{\sin\left(2x\right)}{\sin\left(x\right)}+\frac{-\cos\left(2x\right)}{\cos\left(x\right)}$
2

Using the sine double-angle identity: $\sin\left(2\theta\right)=2\sin\left(\theta\right)\cos\left(\theta\right)$

$\frac{2\sin\left(x\right)\cos\left(x\right)}{\sin\left(x\right)}+\frac{-\cos\left(2x\right)}{\cos\left(x\right)}$
Why does sin(2x) = 2sin(x)cos(x) ?
3

Simplify the fraction $\frac{2\sin\left(x\right)\cos\left(x\right)}{\sin\left(x\right)}$ by $\sin\left(x\right)$

$2\cos\left(x\right)+\frac{-\cos\left(2x\right)}{\cos\left(x\right)}$
4

Combine all terms into a single fraction with $\cos\left(x\right)$ as common denominator

$\frac{2\cos\left(x\right)^2-\cos\left(2x\right)}{\cos\left(x\right)}$
5

Apply the trigonometric identity: $\cos\left(2\theta \right)$$=2\cos\left(\theta \right)^2-1$

$\frac{2\cos\left(x\right)^2-\left(2\cos\left(x\right)^2-1\right)}{\cos\left(x\right)}$
6

Simplify the product $-(2\cos\left(x\right)^2-1)$

$\frac{2\cos\left(x\right)^2-2\cos\left(x\right)^2+1}{\cos\left(x\right)}$
7

Cancel like terms $2\cos\left(x\right)^2$ and $-2\cos\left(x\right)^2$

$\frac{1}{\cos\left(x\right)}$
8

Applying the trigonometric identity: $\displaystyle\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}$

$\sec\left(x\right)$
9

Since we have reached the expression of our goal, we have proven the identity

true

Final answer to the exercise

true

Try other ways to solve this exercise

  • Prove from RHS (right-hand side)
  • Prove from LHS (left-hand side)
  • Express everything into Sine and Cosine
  • Exact Differential Equation
  • Linear Differential Equation
  • Separable Differential Equation
  • Homogeneous Differential Equation
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Load more...
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