Question
Question: How can I solve this sin, cos, tan mathematical equation? \(\sin \left( {{\cos }^{-1}}\left( \dfra...
How can I solve this sin, cos, tan mathematical equation?
sin(cos−1(6563)+2tan−1(43))
Solution
The above trigonometric expression can be solved in 3 steps. Let A = cos−1(6563);
B = 2tan−1(43).In step1, we convert the function A into an inverse tangent function. In step2, we convert the function B into an inverse tangent function. In step3, we simplify the functions in step1 further to get the expression in the form sin(sin−1θ).
Complete step by step solution:
The given trigonometric equation is
sin(cos−1(6563)+2tan−1(43))
Let A = cos−1(6563);
B = 2tan−1(43)
Step1:
We need to convert the function A into inverse tangent function.
From the formulae of trigonometry,
⇒cos−1(ba)=tan−1(ab2−a2)
Applying the same formula for cos−1(6563),
we get,
⇒cos−1(6563)=tan−1(63652−632)
Writing the values of squares,
we get,
⇒cos−1(6563)=tan−1(634225−3969)
⇒cos−1(6563)=tan−1(63256)
We know that 256=16
Substituting the same,
⇒cos−1(6563)=tan−1(6316)
Step2:
We need to convert the function B into inverse tangent function.
From the formula of trigonometry,
⇒2tan−1a=tan−1((1−a2)2a)
Applying the same formula for 2tan−1(43),
we get,
⇒2tan−1(43)=tan−11−(43)22×43
Simplifying the numerator and denominator on the right-hand side,
we get,
⇒2tan−1(43)=tan−11−(169)23
Evaluating the denominator on the right-hand side further,
⇒2tan−1(43)=tan−116(16−9)23
⇒2tan−1(43)=tan−116723
⇒2tan−1(43)=tan−1(23×716)
Cancelling the factors,
⇒2tan−1(43)=tan−1(724)
Step3:
we need to evaluate the two inverse tangent functions in step1 and step2.
The given question is of the form
⇒sin(cos−1(6563)+2tan−1(43))
From step1 and step2,
we can write the above equation as
⇒sin(tan−1(6316)+tan−1(724))
From the formula of trigonometry,
tan−1a+tan−1b=tan−1(1−aba+b)
Applying the same formula for the above expression,
we get,
⇒(tan−1(6316)+tan−1(724))=tan−11−(6316)(724)(6316)+(724)
Taking LCM of the numerator and denominator on right-hand side,
⇒(tan−1(6316)+tan−1(724))=tan−1((63×7)(63×7)−(16×24))((63×7)(16×7)+(24×63))
Simplifying the numerator and denominator on the right-hand side,
we get,
⇒(tan−1(6316)+tan−1(724))=tan−1(441441−384)(441112+1512)
⇒(tan−1(6316)+tan−1(724))=tan−1(44154)(4411624)
Evaluating the above equation further,
⇒(tan−1(6316)+tan−1(724))=tan−1(541624)
Substituting the result in the expression
sin(tan−1(6316)+tan−1(724)),
⇒sin(tan−1(541624))
Now,
From the formula of trigonometry,
⇒tan−1(ba)=sin−1((a2+b2)a)
Applying the same formula for the above expression,
we get,
⇒tan−1(541624)=sin−1((1624)2+(54)2)1624
Writing the values of squares on the right-hand side,
⇒tan−1(541624)=sin−1((2637376)+(2916))1624
Evaluating the expression further,
⇒tan−1(541624)=sin−1((2640292))1624
⇒tan−1(541624)=sin−1(16251624)
Substituting the above value in the expression sin(tan−1(541624)),
we get,
⇒sin(sin−1(16251624))
From the inverse trigonometry,
we know that sin(sin−1θ)=θ
Substituting the same,
⇒sin(sin−1(16251624))=(16251624)
Therefore, the result of the above mathematical equation sin(cos−1(6563)+2tan−1(43)) is (16251624).
Note: We need to know the formulae of trigonometry to solve the problem easily. The above mathematical equation can also be solved by replacing the values of inverse trigonometric functions with an angle.