Short Tricks on Trigonometric Identities

Today we will be covering a very important topic from the Advance Maths part of the Quantitative Aptitude section that is – Important Notes & Short Tricks on Trigonometric Identities.

Important Short Tricks on Trigonometric Identities

Pythagorean Identities

• sin² θ + cos² θ = 1
• tan² θ + 1 = sec² θ
• cot² θ + 1 = csc² θ

Negative of a Function

• sin (–x) = –sin x
• cos (–x) = cos x
• tan (–x) = –tan x
• csc (–x) = –csc x
• sec (–x) = sec x
• cot (–x) = –cot x

If A + B = 90^o, Then

• Sin A = Cos B
• Sin²A + Sin²B = Cos²A + Cos²B = 1
• Tan A = Cot B
• Sec A = Csc B

For example:    

If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?

Solution:            

Tan A = Cot B, Tan A*Tan B = 1

So, A +B = 90^o

(x+y)+(x-y) = 90^o, 2x = 90^o , x = 45^o

Tan (2x/3) = tan 30^o = 1/√3

If A – B = 90^o, (A › B) Then

• Sin A = Cos B
• Cos A = – Sin B
• Tan A = – Cot B

If A ± B = 180^o, then

• Sin A = Sin B
• Cos A = – Cos B

If A + B = 180^o                   

Then, tan A = – tan B

If A – B = 180^o                    

Then, tan A = tan B

For example:    

Find the Value of tan 80^o + tan 100^o ?

Solution:Since 80 + 100 = 180

Therefore, tan 80^o + tan 100^o = 1

If A + B + C = 180^o, then

Tan A + Tan B +Tan C = Tan A * Tan B *Tan C

sin θ * sin 2θ * sin 4θ = ¼ sin 3θ

cos θ * cos 2θ * cos 4θ = ¼ cos 3θ

For Example:What is the value of cos 20^o cos 40^o cos 60^o cos 80^o?

Solution: We know cos θ * cos 2θ * cos 4θ = ¼ cos 3θ

Now, (cos 20^o cos 40^o cos 80^o ) cos 60^o

¼ (Cos 3*20) * cos 60^o

¼ Cos² 60^o = ¼ * (½)² = 1/16

If             a sin θ + b cos θ = m     &    a cos θ – b sin θ = n

then a² + b² = m² + n²

For Example:

If 4 sin θ + 3 cos θ = 2 , then find the value of  4 cos θ – 3 sin θ:

Solution:

Let 2 cos θ – 3 sin θ = x

By using formulae a² + b² = m² + n²

4² + 3² = 2² + x²

16 + 9 = 4 + x²

X = √21

If

sin θ +  cos θ = p     &     csc θ –  sec θ = q

then P – (1/p) = 2/q

For Example:

If sin θ + cos θ = 2 , then find the value of  csc θ – sec θ:

Solution:

By using formulae:

P – (1/p) = 2/q

2-(1/2) = 3/2 = 2/q

Q = 4/3 or csc θ – sec θ = 4/3

If

a cot θ + b csc θ = m     &    a csc θ + b cot θ = n

then b² – a²  = m² – n²

If

cot θ + cos θ = x     &    cot θ – cos θ = y

then x² – y² = 4 √xy

If

tan θ + sin θ = x     &    tan θ – sin θ = y

then x² – y² = 4 √xy

If

y = a² sin²x + b² csc²x + c

y = a² cos²x + b² sec²x + c

y = a² tan²x + b² cot²x + c

then,

y_min = 2ab + c

y_max = not defined

For Example:                    

If y = 9 sin² x + 16 csc² x +4 then y_min is:

Solution:            

For, y min = 2* √9 * √16 + 4

= 2*3*4 + 20 = 24 + 4 = 28

If            

y = a sin x + b cos x + c

y = a tan x + b cot x + c

y = a sec x + b csc x + c

then,     y_min = + [√(a²+b²)] + c

y_max = – [√(a²+b²)] + c

For Example:                    

If y = 1/(12sin x + 5 cos x +20) then y_max is:

Solution:            

For, y max = 1/x min

= 1/- (√12² +5²) +20 = 1/(-13+20) = 1/7

Sin² θ, maxima value = 1, minima value = 0

Cos² θ, maxima value = 1, minima value = 0