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# Prove the trigonometric identity $\sec\left(x\right)=\frac{\sin\left(2x\right)}{\sin\left(x\right)}-\frac{\cos\left(2x\right)}{\cos\left(x\right)}$

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##  Step-by-step Solution 

Problem to solve:

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

Specify the solving method

I. Express the LHS in terms of sine and cosine and simplify

1

Start from the LHS (left-hand side)

$\sec\left(x\right)$
2

Rewrite $\sec\left(x\right)$ in terms of sine and cosine

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

II. Express the RHS in terms of sine and cosine and simplify

3

Start from the RHS (right-hand side)

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

Combine fractions with different denominator using the formula: $\displaystyle\frac{a}{b}+\frac{c}{d}=\frac{a\cdot d + b\cdot c}{b\cdot d}$

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

Apply the trigonometric identity: $\sin\left(x\right)\cos\left(y\right)$$=\frac{\sin\left(x+y\right)+\sin\left(x-y\right)}{2}, where x=2x and y=x \frac{\sin\left(2x+x\right)+\sin\left(2x-x\right)}{2} Combining like terms 2x and x \frac{\sin\left(3x\right)+\sin\left(2x-x\right)}{2} Combining like terms 2x and -x \frac{\sin\left(3x\right)+\sin\left(x\right)}{2} 5 Apply the trigonometric identity: \sin\left(x\right)\cos\left(y\right)$$=\frac{\sin\left(x+y\right)+\sin\left(x-y\right)}{2}$, where $x=2x$ and $y=x$

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

Combine all terms into a single fraction with $2$ as common denominator

$\frac{\frac{\sin\left(3x\right)+\sin\left(x\right)+2\left(-1\right)\cos\left(2x\right)\sin\left(x\right)}{2}}{\sin\left(x\right)\cos\left(x\right)}$

Multiply $2$ times $-1$

$\frac{\frac{\sin\left(3x\right)+\sin\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{2}}{\sin\left(x\right)\cos\left(x\right)}$
6

Combine all terms into a single fraction with $2$ as common denominator

$\frac{\frac{\sin\left(3x\right)+\sin\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{2}}{\sin\left(x\right)\cos\left(x\right)}$
7

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

$\frac{\frac{\sin\left(3x\right)+\sin\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{2}}{\frac{1}{2}\sin\left(2x\right)}$
8

Simplify the fraction $\frac{\frac{\sin\left(3x\right)+\sin\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{2}}{\frac{1}{2}\sin\left(2x\right)}$

$\frac{\frac{1}{\frac{1}{2}}\frac{\sin\left(3x\right)+\sin\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{2}}{\sin\left(2x\right)}$
9

Divide $1$ by $\frac{1}{2}$

$\frac{2\left(\frac{\sin\left(3x\right)+\sin\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{2}\right)}{\sin\left(2x\right)}$
10

Multiplying the fraction by $2$

$\frac{\sin\left(3x\right)+\sin\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{\sin\left(2x\right)}$
11

Apply the trigonometric identity: $\sin\left(a\right)+\sin\left(b\right)$$=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right), where a=3x and b=x \frac{2\sin\left(2x\right)\cos\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{\sin\left(2x\right)} 12 Using the sine double-angle identity: \sin\left(2\theta\right)=2\sin\left(\theta\right)\cos\left(\theta\right) \frac{4\sin\left(x\right)\cos\left(x\right)\cos\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{\sin\left(2x\right)} Why does sin(2x) = 2sin(x)cos(x) ? 13 When multiplying two powers that have the same base (\cos\left(x\right)), you can add the exponents \frac{4\cos\left(x\right)^2\sin\left(x\right)-2\cos\left(2x\right)\sin\left(x\right)}{\sin\left(2x\right)} 14 Factor the polynomial 4\cos\left(x\right)^2\sin\left(x\right)-2\cos\left(2x\right)\sin\left(x\right) by it's greatest common factor (GCF): 2\sin\left(x\right) \frac{2\sin\left(x\right)\left(2\cos\left(x\right)^2-\cos\left(2x\right)\right)}{\sin\left(2x\right)} III. Choose what side of the identity are we going to work on 15 To prove an identity, we usually begin to work on the side of the equality that seems to be more complicated, or the side that is not expressed in terms of sine and cosine. In this problem, we will choose to work on the right side \frac{2\sin\left(x\right)\left(2\cos\left(x\right)^2-\cos\left(2x\right)\right)}{\sin\left(2x\right)} to reach the left side \frac{1}{\cos\left(x\right)} \frac{1}{\cos\left(x\right)}=\frac{2\sin\left(x\right)\left(2\cos\left(x\right)^2-\cos\left(2x\right)\right)}{\sin\left(2x\right)} 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)\left(2\cos\left(x\right)^2-\cos\left(2x\right)\right)}{2\sin\left(x\right)\cos\left(x\right)} Why does sin(2x) = 2sin(x)cos(x) ? Simplify the fraction \frac{2\sin\left(x\right)\left(2\cos\left(x\right)^2-\cos\left(2x\right)\right)}{2\sin\left(x\right)\cos\left(x\right)} by \sin\left(x\right) \frac{2\left(2\cos\left(x\right)^2-\cos\left(2x\right)\right)}{2\cos\left(x\right)} 16 Simplify \frac{2\sin\left(x\right)\left(2\cos\left(x\right)^2-\cos\left(2x\right)\right)}{\sin\left(2x\right)} \frac{2\left(2\cos\left(x\right)^2-\cos\left(2x\right)\right)}{2\cos\left(x\right)} 17 Cancel the fraction's common factor 2 \frac{2\cos\left(x\right)^2-\cos\left(2x\right)}{\cos\left(x\right)} 18 Apply the trigonometric identity: \cos\left(2x\right)$$=2\cos\left(x\right)^2-1$

$\frac{2\cos\left(x\right)^2-\left(2\cos\left(x\right)^2-1\right)}{\cos\left(x\right)}$
19

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

$\frac{2\cos\left(x\right)^2-2\cos\left(x\right)^2+1}{\cos\left(x\right)}$
20

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

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

IV. Check if we arrived at the expression we wanted to prove

21

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

true

true

##  Explore different ways to solve this problem

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