**Trigonometric ratios of compound angles :**

An angle made up of the algebraic sum of two or more angles is called a compound angle.

Here, we are going see the formulas for trigonometric ratios of compound angles.

sin (A + B) = sinA cosB + cosA sinB

sin (A - B) = sinA cosB - cosA sinB

cos (A + B) = cosA cosB - sinA cosB

cos (A - B) = cosA cosB + sinA cosB

tan (A + B) = [tanA + tanB] / [1 - tanA tanB]

tan (A - B) = [tanA - tanB] / [1 + tanA tanB]

From the above table, we can get the values of trigonometric ratios for standard angles such as 0°, 30°, 45°, 60°, 90°

Now, let us look at some practice problems on "Trigonometric ratios of compound angles".

**Example 1 :**

Find the value of cos15°

**Solution :**

First, we have to write the given angle 15° in terms of sum or difference of two standard angles.

So, we have 15° = 45° - 30°

cos15° = cos (45° - 30°)

cos15° = cos45° cos30° + sin45° sin30°

Using the above trigonometric ratio table, we have

cos15° = (√2/2) x (√3/2) + (√2/2) x (1/2)

cos15° = (√6 / 4) + (√2/4)

cos15° = (√6 + √2) / 4

**Hence, the value of cos15° is equal to (√6 + √2) / 4**

**Let us look at the next problem on "Trigonometric ratios of compound angles"**

**Example 2 :**

Find the value of cos105°

**Solution :**

First, we have to write the given angle 105° in terms of sum or difference of two standard angles.

So, we have 105° = 60° + 45°

cos105° = cos (60° + 45°)

cos105° = cos60° cos45° - sin60° sin45°

Using the above trigonometric ratio table, we have

cos105° = (1/2) x (√2/2) - (√3/2) x (√2/2)

cos105° = (√2 / 4) - (√6/4)

cos15° = (√2 - √6) / 4

**Hence, the value of cos15° is equal to (√2 - √6) / 4**

**Let us look at the next problem on "Trigonometric ratios of compound angles"**

**Example 3 :**

Find the value of sin75°

**Solution :**

First, we have to write the given angle 75° in terms of sum or difference of two standard angles.

So, we have 75° = 45° + 30°

sin75° = sin (45° + 30°)

sin75° = sin45° cos30° + cos45° sin30°

Using the above trigonometric ratio table, we have

sin75° = (√2/2) x (√3/2) + (√2/2) x (1/2)

sin75° = (√6 / 4) + (√2/4)

sin75° = (√6 + √2) / 4

**Hence, the value of sin75° is equal to (√6 + √2) / 4**

**Let us look at the next problem on "Trigonometric ratios of compound angles"**

**Example 4 :**

Find the value of tan15°

**Solution :**

First, we have to write the given angle 15° in terms of sum or difference of two standard angles.

So, we have 15° = 45° - 30°

tan15° = tan (45° - 30°)

tan15° = [tan45° - tan30°] / [1 + tan45° tan30°]

Using the above trigonometric ratio table, we have

tan15° = [1 - 1/√3] / [1 + 1x1/√3]

tan15° = [(√3 - 1)/√3] / [(√3 + 1)/√3]

tan15° = [(√3 - 1)/√3] x [(√3/(√3 + 1)]

tan15° = (√3 - 1) / (√3 + 1)

By rationalizing the denominator, we get

tan15° = 2 - √3

**Hence, the value of tan15° is equal to 2 - √3**

Let us look at the next problem on "Trigonometric ratios of compound angles"

**Example 5 :**

If A and B are acute angles, sinA = 3/5, cosB = 12/13,

find cos (A+B)

**Solution :**

cos (A + B) = cosAcosB - sinAsinB

From the formula of cos (A + B), the value of sinA and cosA are given in the question.

We need the value of sinB and cosA.

We know that,

sin ² B = 1 - cos ² B

sin ² B = 1 - (12/13)²

sin ² B = 1 - 144/169

sin ² B = (169 - 144)/169

sin ² B = 25/169

sin ² B = (5/13)² --------> sinB = 5/13

cos ² A = 1 - sin ² A

cos ² A = 1 - (3/5)²

cos ² A = 1 - 9/25

cos ² A = (25 - 9)/25

cos ² A = 16 / 25

cos ² A = (4/5)² --------> cosA = 4 / 5

Plugging the values of cosA, cosB, sinA and sinB in the formula of cos (A+B), we get,

cos (A + B) = (4/5) x (12/13) - (3/5) x (5/13)

cos (A + B) = (4/5) x (12/13) - (3/5) x (5/13)

cos (A + B) = (48 / 65) - (15 / 65)

cos (A + B) = (48 - 15) / 65

cos (A + B) = 33 / 65

**Hence, the value of cos(A+B) is 33/65.**

As we have seen in the above examples, some angles can not be written in terms of sum or difference of two standard angles.

For example,

Let us consider sin 225°

Here, 225° can not be written in terms of sum or difference of two standard angles.

All that we can do is, 225° can be written in terms sum or difference of two angles where one of the angles will be quadrantal angles such as 0°, 90°, 180°, 270°

So, sin 225° can be written as

sin (180° + 45°) or sin (270° - 45°)

To evaluate sin (180° + 45°) or sin (270° - 45°), we have to know ASTC formula

ASTC formla has been explained clearly in the figure given below.

More clearly

In the first quadrant (0° to 90°), all trigonometric ratios are positive.

In the second quadrant (90° to 180°), sin and csc are positive and other trigonometric ratios are negative.

In the third quadrant (180° to 270°), tan and cot are positive and other trigonometric ratios are negative.

In the fourth quadrant (270° to 360°), cos and sec are positive and other trigonometric ratios are negative.

When we have the angles 90° and 270° in the trigonometric ratios in the form of

(90° + θ)

(90° - θ)

(270° + θ)

(270° - θ)

We have to do the following conversions,

sin θ <------> cos θ

tan θ <------> cot θ

csc θ <------> sec θ

For example,

sin (270° + θ) = - cos θ

cos (90° - θ) = sin θ

For the angles 0° or 360° and 180°, we should not make the above conversions.

Now, let us evaluate sin (180° + 45°)

To evaluate sin (90° + θ), we have to consider the following important points.

(i) (180° + 45°) will fall in the III rd quadrant.

(ii) When we have 180°, "sin" will not be changed.

(iii) In the III rd quadrant, the sign of "sin" is negative.

Considering the above points, we have

sin 225° = sin (180° + 45°)

sin 225° = - sin 45°

sin 225° = - √2 / 2

When we are not able to write the given angle in terms of sum or difference of two standard angles, we have to proceed the problem in this way.

**Click here to know more about ASTC formula**

After having gone through the stuff given above, we hope that the students would have understood "Trigonometric ratios of compound angles"

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