If sinq1
+ sinq2
+ sinq3 = 3
then value of cosq1
+ cosq2
+ cosq3 is
Answers
Answer:
Step-by-step explanation:
sin q = 0 Þ q = np,
· cos q = 0 Þq = (2n + 1),
· tan q = 0 Þ q = np,
· sin q = sin a Þq = np + (–1)na, where aÎ [–p/2, p/2]
· cos q = cos aÞq = 2np ± a, where aÎ [ 0, p]
· tan q = tan a Þ q = np + a, where aÎ ( –p/2, p/2)
· sin2 q = sin2 a , cos2 q = cos2 a, tan2q = tan2 aÞq = np±a,
· sin q = 1 Þq = (4n + 1),
· cos q = 1 Þ q = 2np ,
· cos q = –1 Þ q = (2n + 1)p,
· sin q = sin a and cos q = cos aÞ q = 2np + a.
Note:
· Everywhere in this chapter n is taken as an integer, If not stated otherwise.
· The general solution should be given unless the solution is required in a specified interval.
· a is taken as the principal value of the angle. Numerically least angle is called the principal value.
Method for finding principal value
Suppose we have to find the principal value of satisfying the equation sin = – .
Since sin is negative, will be in 3rd or 4th quadrant. We can approach 3rd or 4th quadrant from two directions. If we take anticlockwise direction the numerical value of the angle will be greater than . If we approach it in clockwise direction the angle will be numerically less than . For principal value, we have to take numerically smallest angle.
So for principal value :
1. If the angle is in 1 st or 2nd quadrant we must select anticlockwise direction and if the angle if the angle is in 3rd or 4th quadrant, we must select clockwise direction.
2. Principal value is never numerically greater than .
3. Principal value always lies in the first circle (i.e. in first rotation)
On the above criteria will be or . Among these two has the least numerical value. Hence is the principal value of satisfying the equation sin = –.
Algorithm to find the principle argument:
Step 1: First draw a trigonometric circle and mark the quadrant, in which the angle may lie.
Step 2: Select anticlockwise direction for 1st and 2nd quadrants and select clockwise direction for 3rd and 4th quadrants.
Step 3: Find the angle in the first rotation.
Step 4: Select the numerically least angle among these two values. The angle thus found will be the principal value.
Step 5: In case, two angles one with positive sign and the other with negative sign qualify for the numerically least angle, then it is the convention to select the angle with positive sign as principal value.
Example 1: Iftan = – 1, then will lie in 2nd or 4th quadrant.
For 2nd quadrant we will select anticlockwise and for 4th quadrant. we will select clockwise direction.
In the first circle two values and are obtained.
Among these two, is numerically least angle. Hence principal value is .
Example 2: If cos = , then will lie in 1st or 4th quadrant.
For 1st quadrant, we will select anticlockwise direction and for 4th quadrant, we will select clockwise direction.
In the first circle two values and are thus found.
Both and – have the same numerical value. In such case will be selected as principal value.
Illustration 17: Solve cot (sinx + 3) = 1.
Solution: sinx + 3 = Þ Þ n = 1 Þ sinx =
Þ x = or
Illustration 18: If sin 5x + sin 3x + sin x = 0, then find the value of x other than zero, lying between 0 £ x £.
Solution: sin 5x + sin 3x + sin x = 0 Þ (sin 5x + sin x) + sin 3x = 0
Þ 2 sin 3x cos 2x + sin 3x = 0 Þ sin 3x(2 cos 2x + 1) = 0
Þ sin 3x = 0; cos 2x = – Þ 3x = np, 2x = 2np±
The required value of x is .
Illustration 19: Find all acute angle a such that cos a cos 2a cos 4a = .
Solution: It is given that cosa cos2a cos4a =
Þ 2sina cosa cos2a cos4a = Þ 2sin2a cos2a cos4a =
Þ 2sin4a cos4a = sinaÞ sin8a – sina = 0
Þ 2sincos = 0
Either sin = 0 Þ Þa =
For n = 0 a = 0 which is not a solution.
Þa = n = 1, i.e. a =
or cos Þ = (2n + 1) Þa= (2n + 1) Þa =
Hence a = .
Illustration 20: Solve for x: .
Solution:
Þ
Þ
Þ ÞÞ
Þ sin2x = ± 1 Þ 2x = (2n + 1) Þ x = (2n+1) , n Î I