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

Step-by-step Solution

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Final Answer

true

Step-by-step Solution

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

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Start from the RHS (right-hand side)

$\frac{\sin\left(2x\right)}{\sin\left(x\right)}+\frac{-\cos\left(2x\right)}{\cos\left(x\right)}$
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Combine fractions with different denominator using the formula: $\displaystyle\frac{a}{b}+\frac{c}{d}=\frac{a\cdot d + b\cdot c}{b\cdot d}$

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

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

$\frac{2\sin\left(x\right)\cos\left(x\right)\cos\left(x\right)-\cos\left(2x\right)\sin\left(x\right)}{\sin\left(x\right)\cos\left(x\right)}$
Why does sin(2x) = 2sin(x)cos(x) ?
6

When multiplying two powers that have the same base ($\cos\left(x\right)$), you can add the exponents

$\frac{2\cos\left(x\right)^2\sin\left(x\right)-\cos\left(2x\right)\sin\left(x\right)}{\sin\left(x\right)\cos\left(x\right)}$
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Factor the polynomial $2\cos\left(x\right)^2\sin\left(x\right)-\cos\left(2x\right)\sin\left(x\right)$ by it's greatest common factor (GCF): $\sin\left(x\right)$

$\frac{\sin\left(x\right)\left(2\cos\left(x\right)^2-\cos\left(2x\right)\right)}{\sin\left(x\right)\cos\left(x\right)}$
8

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

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

III. Choose what side of the identity are we going to work on

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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 $\frac{2\cos\left(x\right)^2-\cos\left(2x\right)}{\cos\left(x\right)}$ to reach the left side $\frac{1}{\cos\left(x\right)}$

$\frac{1}{\cos\left(x\right)}=\frac{2\cos\left(x\right)^2-\cos\left(2x\right)}{\cos\left(x\right)}$
10

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

$\frac{2\cos\left(x\right)^2-\left(2\cos\left(x\right)^2-1\right)}{\cos\left(x\right)}$
11

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

$\frac{2\cos\left(x\right)^2-2\cos\left(x\right)^2-1\cdot -1}{\cos\left(x\right)}$
12

Multiply $-1$ times $-1$

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

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

true

Final Answer

true

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Prove from LHS (left-hand side)Prove from RHS (right-hand side)

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Function Plot

Plotting: $true$

Main Topic: Trigonometric Identities

In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables where both sides of the equality are defined.

Used Formulas

1. See formulas

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