Important Short Tricks on Trigonometric Identities
Pythagorean Identities
- sin2 θ + cos2 θ = 1
- tan2 θ + 1 = sec2 θ
- cot2 θ + 1 = cosec2 θ
Negative of a Function
- sin (–x) = –sin x
- cos (–x) = cos x
- tan (–x) = –tan x
- cosec (–x) = –cosec 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 = Cosec B
For example:
If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?
Solution:
If A+B = 90°, then
Tan A = Cot B or 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, and (A › B) Then,
- Sin A = Cos B
- Cos A = - Sin B
- Tan A = - Cot B
- Cosec A = - Sec B
- Sec A = Cosec B
If A ± B = 180o, then
- Cos A = - Cos B
If A + B = 180o
Then,
- Sin A = Sin B
- tan A = - tan B
If A - B = 180o
Then,
- Sin A = - Sin B
- tan A = tan B
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 4 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 & cosec θ + sec θ = q
then p – (1/p) = 2/q
For Example:
If sin θ + cos θ = √2 , then find the value of cosec θ + sec θ:
Solution:
By using formulae:
P – (1/p) = 2/q
√2-(1/√2) = 1/√2 = 2/q
q = 2√2 or cosec θ + sec θ = 2√2
If
a cot θ + b cosec θ = m & a cosec 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 cosec2x + 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 cosec2 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 cosec 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
Here are some important questions of Trigonometric identities.
(1)Value of is
(a)
(b)
(c)
(d)None of these
Ans.(a)
is equal to
(2)If is acute and
then
is equal to
(a)
(b)3
(c) 2
(d) 4
Ans. (c)
If sum of the inversely proportional value is 2
so =2
or we can put = 45°
(3)The simplified value of(Secx Secy + tanx tany)2 - (Secx tany + tanx Secy)2 is
(a)-1
(b)0
(c)sec2x
(d)1
Ans. (d)
The simplified value of (Secx Secy + tanx tany)2 - (Secx tany + tanx Secy)2 is obtained by putting x = y = 45°
= (√2×√2 + 1×1)2 - (√2×1 + 1×√2)2
= (2+1)2 - (√2+√2)2
=(3)2 - (2√2)2
= 9 - 8 = 1
(4) Find the value of
(a) 1
(b) -1
(c) 2
(d) -2
Ans. (c)
put
(5) If then
is equal
(a)7/4
(b) 7/2
(c)5/2
(d)5/4
Ans. (d)
on solving we get sec= 5/4
Note: if x+y=a
and x-y=b
then x=(a+b)/2 and y=(x-y)/2
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