Short Tricks on Trigonometric Identities

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 Notes & Short Tricks on Trigonometric Identities

Important Short Tricks on Trigonometric Identities

Pythagorean Identities

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  • 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 = 90o, 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 = 90o

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

Tan (2x/3) = tan 30o = 1/√3

If A – B = 90o, (A › B) Then

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

If A ± B = 180o, then

  • Sin A = Sin B
  • Cos A = – Cos B

If A + B = 180o                   

Then, tan A = – tan B

If A – B = 180o                    

Then, tan A = tan B

For example:    

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

Solution:Since 80 + 100 = 180

Therefore, tan 80o + tan 100o = 1

If A + B + C = 180o, 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 20o cos 40o cos 60o cos 80o?

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

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

¼ (Cos 3*20) * cos 60o

¼ Cos2 60o = 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,

ymin = 2ab + c

ymax = not defined

For Example:                    

If y = 9 sin2 x + 16 csc2 x +4 then ymin 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,     ymin = + [√(a2+b2)] + c

ymax = – [√(a2+b2)] + c

For Example:                    

If y = 1/(12sin x + 5 cos x +20) then ymax 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

 

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