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source: ncert
topic: trigonometry
class_level: class_12
chapter: ch2
difficulty: advanced
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Class 12 | Ch2: Inverse Trigonometric Functions
DOMAINS AND RANGES (Principal Value Branch):
sin^(-1): domain [-1,1], range [-pi/2, pi/2].
cos^(-1): domain [-1,1], range [0, pi].
tan^(-1): domain R, range (-pi/2, pi/2).
cosec^(-1): domain R-(-1,1), range [-pi/2,pi/2]-{0}.
sec^(-1): domain R-(-1,1), range [0,pi]-{pi/2}.
cot^(-1): domain R, range (0,pi).
KEY PROPERTIES:
sin^(-1)(x)+cos^(-1)(x)=pi/2 for x in [-1,1].
tan^(-1)(x)+cot^(-1)(x)=pi/2 for all x.
sec^(-1)(x)+cosec^(-1)(x)=pi/2 for |x|>=1.
sin^(-1)(-x)=-sin^(-1)(x). cos^(-1)(-x)=pi-cos^(-1)(x).
tan^(-1)(-x)=-tan^(-1)(x).
sin^(-1)(1/x)=cosec^(-1)(x) for |x|>=1.
cos^(-1)(1/x)=sec^(-1)(x) for |x|>=1.
tan^(-1)(1/x)=cot^(-1)(x) for x>0.
ADDITION FORMULAS:
tan^(-1)(x)+tan^(-1)(y)=tan^(-1)((x+y)/(1-xy)) if xy<1.
tan^(-1)(x)+tan^(-1)(y)=pi+tan^(-1)((x+y)/(1-xy)) if xy>1 and x>0.
2 tan^(-1)(x)=sin^(-1)(2x/(1+x^2))=cos^(-1)((1-x^2)/(1+x^2))=tan^(-1)(2x/(1-x^2)).
SOLVED EXAMPLES:
Example 1: Find principal value of sin^(-1)(-1/2).
sin^(-1)(1/2)=pi/6. sin^(-1)(-1/2)=-pi/6. (In range [-pi/2,pi/2]).
Example 2: Find principal value of cos^(-1)(-sqrt3/2).
cos^(-1)(sqrt3/2)=pi/6. cos^(-1)(-sqrt3/2)=pi-pi/6=5pi/6. (In range [0,pi]).
Example 3: Simplify tan^(-1)(cosx-sinx)/(cosx+sinx).
Divide by cosx: tan^(-1)((1-tanx)/(1+tanx))=tan^(-1)(tan(pi/4-x))=pi/4-x.
(Valid when pi/4-x is in range (-pi/2,pi/2).)
Example 4: Prove sin^(-1)x+cos^(-1)x=pi/2.
Let sin^(-1)x=theta. Then x=sin(theta). cos^(-1)x=cos^(-1)(sin theta)=cos^(-1)(cos(pi/2-theta))=pi/2-theta.
Sum=theta+pi/2-theta=pi/2. Proved.
COMMON MISTAKES:
sin^(-1)(x) is NOT 1/sin(x). It is the inverse FUNCTION.
Principal value of cos^(-1): range is [0,pi], NOT [-pi/2,pi/2].
cos^(-1)(-x)=pi-cos^(-1)(x) (NOT -cos^(-1)(x) like sine). |