How to prove trignometric identities: A comprehensive Guide
- Author: Noreen Niazi
- Last Updated on: February 5, 2024
How to prove trigonometric identities? Is there’s any shortcut to prove trigonometric identities? Lets dive into easiest method to prove trigonometric identities?
Either you are students of 9th grade or a teacher, you often come through the trigonometric identities. Trigonometric identities are the identities that use the properties of right triangle and when put values in it, it will always true.
What are trigonometric identities?
Trigonometric identities are the equations that are true for all values of variables.
For example, consider the $$cos^2 \theta +sin^2 \theta =1 $$. Now here you can take any value $$\theta$$, result is true for all values.
Let $$\theta=90$$ then
$$cos^2 \theta +sin^2 \theta =1$$
$$\theta=90$$. Let’s first solve the left-hand side
$$L.H.S=cos^2 \theta +sin^2 \theta =1$$
$$=cos^2 90 +sin^2 90 $$
$$cos 90=0 \text{and} sin90=1$$
$$=0+1$$
$$=1$$
Here left-hand side is equal to the right-hand side.
How to prove trigonometric identities?
We solve trigonometric identities by showing that a specified identity holds for values of (x) or $$(\theta ) $$and in fact, is true for all such essentials.
- Start with One Side: First, solve processes each side separately. You’ll manipulate it by utilizing algebraic and trigonometric identities for its simplification.
- Use Known Identities: Use other held trigonometric identities; some of the Pythagorean identities or sum / product formulas. In the same regard, lets think of an identity $$(sin^2 \theta + cos^2 \theta = 1$$).
- Factor and Simplify: Identify chances to simplify by using factor expressions, square binomials, or term conversion in fractions.
- Substitute and Transform: the ones which appear in the final expression are note. All suspensive known identities should be replaced and correct substitutes existing to fulfill the conditions of validity should apply here. Alternatively, alter terms into sines and cosines.
Examples
Prove that $$ (1 – \sin x)(1 + \csc x)=\frac{\cos^2 x}{\sin x}$$
We have:
$$ (1 – \sin x)(1 + \csc x)$$
$$= (1 – \sin x)(1 + \frac{1}{\sin x})$$
Expanding the brackets:
$$ = 1 + \frac{1}{\sin x} – \sin x – 1 $$
$$= \frac{1}{\sin x} – \sin x $$
$$= \frac{1 – \sin^2 x}{\sin x} $$
$$=\frac{\cos^2 x}{\sin x}$$
$$= \cos x \cot x $$
What are the standard trigonometric formulas and identities?
Standard trigonometry formulas or identities are the set of equations that linked different trigonometric function with each other. Trigonometric formulas also linked different functions which each other. There are six standards trigonometric. These are
- Sine function $$\sin (x)$$
- Cosine function $$\cos (x)$$
- Tangent function $$ \tan (x) $$
- Cosecant function $$\csc (x)$$
- Secant function $$\sec (x)$$
- Cotangent function $$\cot(x) $$
Trigonometric functions values can be determined with the help of right angles triangles. Lets discuss how we can find the values of trigonometric ratios with help of right-angle. triangles.
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Trigonometric ratios:
Since there are six trigonometric ratios. Let’s prove them one by one. Now since the value of sine can be find as ratio of perpendicular and hypotheses. Similarly, we can also find the value of other identities with the help of right angle triangles.
- Sine function $$\sin (x)=\frac { \text{Opposite side}}{\text {Hypotheous sides}}$$
- Cosine function $$\cos (x)=\frac { \text{Adjacent side}}{\text {Hyotheneus}}$$
- Tangent function $$ \tan (x) =\frac { \text{Opposite side}}{\text {Adjacent}}$$
- Cosecant function $$\csc (x)=\frac { \text{Hyothenous side}}{\text {Opposite}}$$
- Secant function $$\sec (x)=\frac { \text{Hypthenous side}}{\text {Adjacent side}}$$
- Cotangent function $$\cot(x) = \frac{\text{Adjacent}}{\text{Opposite side}}$$
Solved Examples of Trignometric Ratios?
Example 1:
- Problem: Find the value of $$( \tan 30^\circ + \sin 60^\circ )$$.
- Solution:
- We know that $$( \tan 30^\circ = \frac{1}{\sqrt{3}} )$$ and $$( \sin 60^\circ = \frac{\sqrt{3}}{2} ).$$
- Adding both values: $$[ \tan 30^\circ + \sin 60^\circ = \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} ]$$
- Rationalizing the denominator: $$[ \frac{2 + \sqrt{3} \cdot \sqrt{3}}{2\sqrt{3}} = \frac{2 + 3}{2\sqrt{3}} = \frac{5}{2\sqrt{3}} ]$$
Example 2:
- Problem: Calculate $$( \sin 45^\circ – \cos 45^\circ ).$$
- Solution:
- We have $$( \sin 45^\circ = \frac{1}{\sqrt{2}} )$$ and $$( \cos 45^\circ = \frac{1}{\sqrt{2}} ).$$
- Subtracting: $$[ \sin 45^\circ – \cos 45^\circ = \frac{1}{\sqrt{2}} – \frac{1}{\sqrt{2}} = 0 ]$$
Example 3: Escalator Stairs:
- Problem: A person is standing on an escalator moving upward. The angle of inclination with the ground is 30°. If the person travels 76 feet on the escalator stairs, find the height gained.
- Solution:
- We have the angle $$( \theta = 30^\circ ).$$
- The trigonometric ratio that involves the opposite side and hypotenuse is $$( \sin \theta ).$$
- Using the given information: $$[ \sin 30^\circ = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{76}{d} ]$$
- Solving for $$( d ): [ d \cdot \sin 30^\circ = 76 ] [ d = \frac{76}{\sin 30^\circ} ] [ d = 152 ]$$
- So, the person gains a height of 152 feet on the escalator stairs.
Example 4: Building Shadow:
- Problem: A building casts a shadow of 60 meters when the angle of elevation of the sun is 45°. Find the height of the building.
- Solution:
- We have the angle of elevation $$( \theta = 45^\circ )$$.
- The trigonometric ratio that involves the opposite side and adjacent side is $$( \tan \theta )$$.
- Using the given information: $$[ \tan 45^\circ = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{h}{60} ]$$
- Solving for the height ( h ): $$[ h = 60 \cdot \tan 45^\circ = 60 \cdot 1 = 60 ]$$
- The height of the building is 60 meters
Pythagorean Trigonometric Identities
Derived from the Pythagorean theorem:
- $$(\sin^2 \theta + \cos^2 \theta = 1)$$
- $$(\sec^2 \theta – \tan^2 \theta = 1)$$
- $$(\csc^2 \theta – \cot^2 \theta = 1)$$
Proof for the first Pythagorean identity: Consider a right-angled triangle ABC with a right angle at B:
- Apply the Pythagoras theorem: $$\text{Opposite}^2 + \text{Adjacent}^2 = \text{Hypotenuse}^2$$
- Divide both sides by the hypotenuse squared: $$\frac{\text{Opposite}^2}{\text{Hypotenuse}^2} + \frac{\text{Adjacent}^2}{\text{Hypotenuse}^2} = 1$$
- Simplify to get: $$\sin^2 \theta + \cos^2 \theta = 1$$
Proof for the second Pythagorean identity: Consider a right-angled triangle ABC with a right angle at B:
- Apply the Pythagoras theorem: $$\text{Opposite}^2 + \text{Adjacent}^2 = \text{Hypotenuse}^2$$
- Divide both sides by the opposite squared: $$\frac{\text{Opposite}^2}{\text{Opposite}^2} + \frac{\text{Adjacent}^2}{\text{Opposite}^2}=\frac{\text{Hypothenous}^2}{\text{Opposite}^2}$$
- Cotangent function $$\cot (x)=\frac { \text{Adjacent side}}{\text {Opposite}}$$
- Cosecant function $$\csc (x)=\frac { \text{Hyothenous side}}{\text {Opposite}}$$
- Simplify to get: $$1 + \cot^2 \theta = \csc^2 \theta$$
Proof for the third Pythagorean identity: Consider a right-angled triangle ABC with a right angle at B:
- Apply the Pythagoras theorem: $$\text{Opposite}^2 + \text{Adjacent}^2 = \text{Hypotenuse}^2$$
- Divide both sides by the adjacent squared: $$\frac{\text{Opposite}^2}{\text{Adjacent side}^2} + \frac{\text{Adajcent}^2}{\text{Adjacent}^2}=\frac{\text{Hypothenous}^2}{\text{Adjacent}^2}$$
- Tangent function $$ \tan (x) =\frac { \text{Opposite side}}{\text {Adjacent}}$$
- Secant function $$\sec (x)=\frac { \text{Hypthenous side}}{\text {Adjacent side}}$$
- Simplify to get: $$1 + \tan^2 \theta = \sec^2 \theta$$
Conclusion:
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