Question

someone help me please.....

Answer 1

I think that this question is not complite

Answer 2

Answer:

ANSWERS ARE DIFFERENT FOR DIFFERENT RATIOS

Step-by-step explanation:

Find the values of trigonometric ratios :  

1) sin (315)  

Solution : sin (315)0

sin (315)0 = sin (90 x 3 + 45)  

Since 315 lies in the 4th quadrant and in this quadrant sine function is negative , also 3 is an odd integer.  

∴ sin (315)0 = sin (90 x 3 + 45) = - cos 45 = −12√

2) cos (210)  

Solution : cos (210)0

cos (210)0 = cos(90 x 2 + 30)  

Since 210 lies in the 3rd quadrant and in this quadrant cosine function is negative.Also the multiple of 90 is even.  

∴ cos (210)0 = cos (90 x 2 + 30) = - cos 30 = −3√2

3) cos (480)  

Solution : cos (480)0

cos (480)0 = cos(90 x 5 + 30)  

Since 480 lies in the 2nd quadrant and in this quadrant cosine function is negative.Also the multiple of 90 is odd.  

∴ cos (480)0 = cos (90 x 5 + 30) = -sin 30 = −12

4) csc (390)  

Solution : csc (390)0

csc (390)0 = csc (90 x 4 + 30)  

Since 390 lies in the 1st quadrant and in this quadrant sine function is positive so cosecant function is also positive , also 4 is an even integer.  

∴ csc (390)0 = csc (90 x 4 + 30) = csc 30 = 2  

5) tan 19π3  

Solution : tan 19π3  

19π3=(193×180) = 1140  

tan (1140)0 = tan (90 x 12 + 60)  

Since 1140 lies in the 1st quadrant and in this quadrant tangent function is positive , also 12 is an even integer.  

∴ tan 19π3= tan (90 x 12 + 60) = tan 60 = 3–√  

6) Prove that cos (510)0 cos (330)0 + sin (390)0 cos (120)0 = -1

Solution :  

We have,  

cos (510)0 cos (330)0 + sin (390)0 cos (120)0  

= cos(90 x 5 + 60) cos(90 x 3 + 60) + sin(90 x 3 + 30) cos(90 x 1 + 30)

= (- sin 60) (sin 60) + (sin 30)(- sin 30)

= −3√2×3√2+12×−12

= −34−14  

= -1  

∴ cos (510)0 cos (330)0 + sin (390)0 cos (120)0 = -1