Trigonometric Ratios of Submultiple of an Angle | IIT JEE Ratios of Submultiple of an Angle

If A is an angle, then 2A, 3A, 4A etc. are called ‘multiple’ angles of A and A/2, A/3 etc. are called ‘sub multiple’ angles of A. Trigonometry is flooded with the formulae of multiple and submultiple of an angle and these formulae are extremely important as they often fetch some direct questions in the exam.

The major formulae for trigonometric ratios of submultiple of an angle are listed below:

• | sin A/2 + cos A/2 | = √(1 + sin A)

• sin A/2 + cos A/2 =  + √(1 + sin A), if 2nπ – π/4 ≤ A/2 ≤ 2nπ + 3π/4

 = - √(1 + sin A), otherwise

• | sin A/2 - cos A/2 | = √(1 - sin A)

• sin A/2 - cos A/2 =  + √(1 - sin A), if 2nπ + π/4 ≤ A/2 ≤ 2nπ + 5π/4

= - √(1 - sin A), otherwise

• tan A/2 = ± √(1-cos A) / (1+cos A)

• a cos A + b sin A < √a² + b²

The ambiguities of signs can be removed according to the quadrants in which the function lies.
 

Watch this Video for more reference

Important points to be noted

• The formula that gives the value of sin A/2 in terms of sin A shall also give the value of sin (nπ + (-1)n) A / 2.

• On parallel lines, the formula applicable for cos A/2 shall also hold good for cos of 2nπ ± A/2.

• Similarly, the formula giving the value of tan A/2 in terms of tan A shall also give the value of tan of nπ +A/2.

• Some results that hold good when A + B + C = π are:

1. sin² A/2 + sin²  B/2 + sin²  C/2  = 1 – 2 sin A/2 sin B/2 sin C/2

2. cos² A/2 + cos² B/2 + cos² C/2  = 2 + 2 sin A/2 sin B/2 sin C/2

3. sin² A/2 + sin² B/2 - sin2 C/2  = 1 – 2 cos A/2 cos B/2 sin C/2

4. cos² A/2 + cos² B/2 - cos² C/2  = 2 + 2 cos A/2 cos B/2 sin C/2

5. tan B/2 tan C/2 + tan C/2 tan A/2 + tan A/2 tan B/2 =1

6. cot A/2 cot B/2 cot C/2 = cot A/2 + cot B/2 + cot C/2  
    

Illustration:

Find the values of sin 67 1°/2 and cos 67 1°/2 .

Solution:

Let us assume A = 67 1°/2.

Then we know the result cos 2A = 2cos2A – 1 =1 - 2sin2A

If we consider A = 67 1°/2 in the above result then we have

cos 135° = 2cos2(67.5°) -1 and since cos135° = - cos45° = -1/√2

Hence, -1/√2 = 2cos2(67.5°) -1 so

2cos2(67.5°) = 1 - 1/√2  

=1 - √2/2

=(2 - √2)/2

Hence, cos2(67.5°) = (2 - √2) /4

cos (67.5°) = √(2-√2) /2

Note that we haven’t considered the negative value as cos (67.5°) is positive.

 For sin 67.5°, we use the formula cos 2A=1 - 2sin² A

Using A = 67.5° and proceeding on the same lines we get the value of sin 67.5° as √(2 + √2) /2.

Illustration:

If A, B and C are the angles of a triangle, then show that

tan² A/2 + tan² B/2 + tan² C/2 ≥ 1

Solution:

We have (A/2 + B/2 + C/2) = π/2, so that

tan (A/2 + B/2)  = tan (π/2 - C/2) 

⇒ (tan A/2 + tan B/2) / (1 - tan A/2 tan B/2) = 1/ (tan C/2)

⇒ tan A/2  tan B/2  + tan B/2  tan C/2  + tan C/2  tan A/2   = 1

⇒ tan² A/2 + tan² B/2 + tan² C/2 – 1 = 1/2 [ ∑2 tan² A/2 - ∑ 2 tan A/2 tan B/2] .

= 1/2 [(tan A/2 - tan B/2)² + (tan B/2 - tan C/2)² + (tan C/2 - tan A/2)²] > 0

 Hence, tan² A/2 + tan² B/2 + tan² C/2 > 1

Illustration:

The equation 2 cos² (x/2) sin 2x = x² + x^-2, x ≤ π/9 has

(1) no real solution

(2) one real solution

(3) more than one real solution

(4) None of the above

Solution:

The given equation is 2 cos² (x/2) sin 2x = x² + x^-2, x ≤ π/9

Since cos and sine lie between -1 and 1 hence,

the L.H.S i.e. 2 cos² (x/2) sin 2x < 2

But the R.H.S i.e. x² + x^-2 = x² + 1/x² ≥ 2.

Hence the equation has no real roots and so the correct option is (1).
 

Related Resources

• Look into the Revision Notes on Trigonometry for a quick revision.

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