If 2cos^2@=1+2sin^2@ and 0<@<π/2,@=?
Answers
Answer:
Step-by-step explanation:
In the earlier sections we have already learnt how to solve trigonometric equations in one variable. But you must have seen various questions in the past year papers involving more than one variable. There are some standard patterns of questions on trigonometric equations with more than one variable which are always asked in most of the competitive exams.
Once you are comfortable and have a solid foundation of trigonometric equations with one variable, this topic won’t be very difficult for you. Though there is no established way of dealing with all the problems but once you go through various possible ways, you will surely be able to handle the IIT JEE problems.
Kindly go through the below listed tricks and the standard steps of dealing with such problems. Most of the problems can be tackled by these tips:
(1) If possible, reduce the equation in terms of any one variable, preferably x. Then solve the equation as you used to in case of a single variable.
(2) Try to derive the linear/algebraic simultaneous equations from the given trigonometric equations and solve them as algebraic simultaneous equations.
(3) At times, you might be required to make certain substitutions. It would be beneficial when the system has only two trigonometric functions.
Illustration: Solve for general x, y,
sin (x – y) = 2 sin x sin y, where x and y are two acute angles of right angled triangle.
Solution: Since it is given that x and y are two acute angles of right triangle, so we make the substitution y = π/2 – x in first equation and transform it as follows:-
sin (2x - π/2) = 2 sin x sin (π/2 - x)
⇒ – sin (π/2 - 2x) = 2sin x cos x
⇒ –cos 2x = sin 2x
⇒ tan 2x = –1 = tan (-π/4)
⇒ 2x = n π – π/4, where n = 0, ±1, ±2, ………
or, x = nπ/2 - π/8, where n = 0, ±1, ±2, ………
y = nπ/2 - 5π/8, where n = 0, ±1, ±2, ………
Hence solution is {nπ/2 + π/8, 5π/8 - πn/2}.
Illustration: Given that sin x (cos y + 2sin y) – cos x (2cos y - sin y) = 0, find the value of tan (x + y).
Solution: sin x (cos y + 2sin y) – cos x (2cos y - sin y) = 0
Remove the parentheses:
sin x*cos y + 2sin x*sin y - 2cos x*cos y + cos x*sin y = 0
Rearrange the terms:
sin x*cos y + cos x*sin y - 2cos x*cos y + 2sin x*sin y = 0
Factor -2 out of the 3rd and 4th terms:
sin x*cos y + cos x*sin y - 2(cos x*cos y – sin x*sin y) = 0
Use the identity sin (A + B) = sin A*cos B + cos A*sin B to rewrite the first two terms:
sin (x + y) - 2(cos x*cos y – sin x*sin y) = 0
Use the identity cos(A + B) = cos A*cos B - sinA*sin B to rewrite the two terms inside the parentheses. Hence, the equation gets reduced to
sin (x + y) - 2*cos(x + y) = 0
Divide throughout by cos(x + y)
{sin (x + y)}/ {cos (x + y)} – {2*cos (x + y)} / {cos (x+y)} = 0
Using the identity tan A = sin A/ cos A to rewrite the beginning terms and after simplification, we obtain
tan (x + y) - 2 = 0
which gives tan (x + y) = 2.
Watch this Video for more reference
Illustration: Solve the system of equations
sin2 x + sin2 y = 1/2.
x – y = 4π/3
Solution: We transform the first equation of the system–
1/2 (1 – cos 2x) + 1/2 (1 – cos 2y) = ½
cos 2x + cos 2y = 1 &
2 cos (x + y).cos(x – y) = 1
Hence it is clear that the system–
cos (x + y) cos (x – y) = 1/2 …… (1)
x – y = 4π/3 …… (2)
has the same solution as the original system, i.e. the systems are equivalent.
So, from equation (1) and (2) we have
cos (x + y) cos (4π/3) = 1/2 or cos (x + y) = –1
x + y = 2nπ ± π
Hence we have two linear equations in x and y
x + y = 2n π ± π, n ∈ I
x – y = 4π/3
x = nπ + 2π/3 ± π/2 …… (a)
Taking +ve sign of (a)
x = (n +7/6)π
= kπ, k = (n +7/6) and y = kπ – 4π/3
Taking –ve sign of (a)
x = nπ + π/6
y = nπ + π/6-4π/3
= n π – 7π/6.
The general solution of system of equations is given by (kπ, kπ - 4π/3) and (nπ + π/6, nπ -7π/2) where k = n + 7/6, n ∈ I
So, these are the solutions to the original system.
Note: In the previous examples we have written the relationships between the unknowns x and the set of solution of the system has been expressed in terms of only one integral parameter e.g. n, k etc. But in practical applications, we may sometimes require to express the general solution in terms of two integral parameters while solving a system of equations with two variables.