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MAPUA UNIVERSITY
INTRAMUROS, MANILA
CORRELATION – 1
TRIGONOMETRY
A. Graph and Properties of Trigonometric Functions
1.) What quadrant where the sine and cosine functions are both increasing?
Trigonometric
Function
Quadrant Range Increasing?
Sine
I [
] Yes
II [ 1 , 0 ] No
III
[
]
No
IV [− 1 , 0 ] Yes
Cosine
I [ 1 , 0 ] No
II [
] No
III [− 1 , 0 ] Yes
IV
[
]
Yes
Thus, at Quadrant IV, sine and cosine functions are both increasing.
2.) Find the amplitude, period and the frequency of the curve, y=2sin(x/2).
Recall:
Model Equation: 𝑦 = 𝑎𝑠𝑖𝑛(𝑏𝑥)
Where:
𝑎 is the amplitude.
Period (𝑇) =
Frequency (𝑓) =
Solution:
Since 𝑎 = 2 and 𝑏 =
Thus,
Amplitude
Period
1
2
Frequency
3.) Find the period of the curve, y=tan 3x.
Solution:
Since 𝑏 = 3 and the function is tangent. Thus,
Period
𝜋
⌈𝑏⌉
4.) Find the amplitude of the sinusoidal curve y= 3cos2x + 4sin2x.
Solution:
𝑎 = 3 and 𝑏 = 4
Solving for 𝐴:
𝐴 = − sin
− 1
The value of A is negative since it lies in the third quadrant.
At Quadrant II:
sin 𝐵 =
Solving for B:
𝐵 = sin
− 1
The value of B is positive, since it lies in the second quadrant.
Thus, sin(𝐴 + 𝐵):
sin ([− sin
− 1
)] + [sin
− 1
)]) =
3.) Simplify the expression 4 𝑐𝑜𝑠𝑦𝑠𝑖𝑛𝑦[ 1 − 2 𝑠𝑖𝑛
2
(𝑦)]
4 cos(𝑦) sin(𝑦)[ 1 − 2 𝑠𝑖𝑛
2
(𝑦)]
The expression can also be expressed as:
●
2 sin(𝑦) cos(𝑦)[ 1 − 2 sin( 2 𝑦)]
Recall:
sin
= 2 sin
cos
cos( 2 𝑦) = 1 − 2 sin
2
Solution:
Thus, Applying the Double Angle Formula:
[
2 sin
cos
)][
1 − 2 sin
)]
2 [sin( 2 𝑦)][cos( 2 𝑦)]
Then, Applying the Double Angle Formula:
sin
= 2 sin
cos
[
sin
)][
cos
)]
4.) If 𝑥𝑐𝑜𝑠𝜃 + 𝑦𝑠𝑖𝑛𝜃 = 1 and 𝑥𝑠𝑖𝑛𝜃 – 𝑦𝑐𝑜𝑠𝜃 = 3 , find the relationship between 𝑥 and 𝑦.
Solution:
Square both sides:
2
2
2
(cos 𝜃)
2
2
(sin 𝜃)
2
2
2
2
sin 𝜃
2
− 2 𝑥𝑦 cos 𝜃 sin 𝜃 + 𝑦
2
cos 𝜃
2
Then add:
[𝑥
2
(cos 𝜃)
2
2
(sin 𝜃)
2
= 1 ]
[
2
sin 𝜃
2
− 2 𝑥𝑦 cos 𝜃 sin 𝜃 + 𝑦
2
cos 𝜃
2
]
[ 𝑥
2
( cos 𝜃
)
2
2
( sin 𝜃
)
2
]
[ 𝑥
2
( sin 𝜃
)
2
− 2 𝑥𝑦 cos 𝜃 sin 𝜃 + 𝑦
2
( cos 𝜃
)
2
] = 1 + 9
2
(cos 𝜃)
2
2
(sin 𝜃)
2
2
(sin 𝜃)
2
2
(cos 𝜃)
2
2
(cos 𝜃)
2
2
(sin 𝜃)
2
2
(sin 𝜃)
2
2
(cos 𝜃)
2
2
[(
cos 𝜃
2
sin 𝜃
2
]
2
[(
sin 𝜃
2
cos 𝜃
2
]
Note:
cos 𝜃
2
sin 𝜃
2
Thus:
2 sin
2
2
cos(𝐴) =
Solution:
sin
cos(𝐴) =
Thus,
A
7.) In what quadrant will angle A terminates if sec A is positive and csc A is negative?
Recall:
sec 𝐴 =
cos 𝐴
csc 𝐴 =
sin 𝐴
Trigonometric
Function
Quadrant Range Value?
Positive (+) or
Negative (−)
I [ 0 , 1 ] +
II
[
]
Sine III
[
]
IV [− 1 , 0 ] −
Cosine
I [ 1 , 0 ] +
II [
]
III [− 1 , 0 ] −
IV
[
]
Since, the value of cosine is positive and the value of sine is negative at Quadrant IV.
Therefore, at Quadrant IV, the value of secant is positive, and the value of cosecant is negative.
8.) If sin A = 3/5 and A is in the second quadrant, while cos b = 7/25 and B is in the first
quadrant, find sin (A + B).
Solution:
At Quadrant II:
sin 𝐴 =
Solving for 𝐴:
𝐴 = sin
− 1
The value of A is positive since it lies in the third quadrant.
At Quadrant I:
cos 𝐵 =
Solving for B:
𝐵 = cos
− 1
The value of B is positive since it lies in the first quadrant.
Thus, sin(𝐴 + 𝐵):
sin 𝑥
= csc 𝑥
Therefore:
1 + sec 𝑥
tan 𝑥 + sin 𝑥
10.) The terminal side of - 1500° will lie in what quadrant:
Note:
Since the angle is negative, it indicates that the rotation is clockwise with respect to x – axis.
Recall:
Given:
Let 𝜃 be the −1500°
Solution:
At 𝑛 = 1 ,
At 𝑛 = 2 ,
At 𝑛 = 3 ,
At 𝑛 = 4 ,
Since at 4
th
Iteration, the angle is −60°. Thus, it implies that the terminal side of −1500° lies at
Quadrant IV.
11.) If tan
− 1
𝑥 + tan
− 1
1
3
) = 45°, find 𝑥.
csc 𝑥
Solution:
tan
− 1
𝑥 + tan
− 1
Solving for 𝑥:
𝑥 = tan ( 45 ° − tan
− 1
1 2.) If the product of 𝑐𝑠𝑐(
𝑥
2
) and 𝑐𝑜𝑠(
𝑥
3
- 60°) is equal to 1. Find the value of x.
Solution:
csc (
) cos (
Using the Reciprocal Identity:
csc (
sin (
Thus,
sin (
cos (
Multiplying both sides by sin (
𝑥
2
cos (
Using Cofunction Identity:
sin 𝑥 = cos ( 90 ° − 𝜃)
In this case, the expression:
𝑥
3
- 60 ° will be the 𝜃. Thus,
Simplifying:
Solving for 𝑥:
14.) If sin A + sin B = 1 and sin A – sin B = 1, find A.
Solution:
Add both equation:
[
sin 𝐴 + sin 𝐵
]
[
sin 𝐴 − sin 𝐵
]
2 sin 𝐴 = 2
Solving for 𝐴:
𝐴 = sin
− 1
15.) Simplify
sin 2 𝑥
1 +cos 2 𝑥
Recall:
sin
= 2 sin
cos
cos( 2 𝑥) = 1 − 2 𝑠𝑖𝑛
2
Solution:
sin 2 𝑥
1 + cos 2 𝑥
2 sin 𝑥 cos 𝑥
2
Simplify:
2 sin 𝑥 cos 𝑥
2
2 sin 𝑥 cos 𝑥
2
2 sin 𝑥 cos 𝑥
2
sin 𝑥 cos 𝑥
2
Using the Pythagorean Identity: 𝑐𝑜𝑠
2
2
sin 𝑥 cos 𝑥
2
sin 𝑥 cos 𝑥
2
sin 𝑥
cos 𝑥
= tan 𝑥
Thus,
sin 2 𝑥
1 + cos 2 𝑥
C. Problems dealing with Right Triangles
1.) Five small circular holes are punched evenly on the rim of circular plate with diameter of
20cm. Find the distance between two adjacent holes.
tan 𝑥
Since a circle is equivalent to 360°, and there are 5 small circle on the rim, Thus,
Since the right triangle, has a hypothenuse with a length of 10 cm and an opposite side with a
length equal to L/2.
sin (
sin( 36 ) =
Solving for 𝐿:
𝐿 = 20 sin( 36 ) ≈
2.) A transmitter with a height of 15m is located on top of the mountain which 3km high. What is
the farthest distance on the surface of the earth that can be seen from the top of the mountain?
Take the radius of the earth to be 6400km.
Let ℎ be the height of transmitter on top of the mountain in kilometers (km).
Let 𝑅 be the radius of the Earth given as 6400 km.
Let 𝑡 be the height of transmitter given as 15 meters.
Let 𝑚 be the height of the mountain given as 3 km.
Let 𝑑 be the farthest distance on the surface of the earth that can be seen from the top of the
mountain.
Solution:
Converting the height of transmitter in terms of kilometers:
𝑡 = 15 𝑚 ×
Thus, the height of transmitter on top of the mountain plus the height of the mountain will be:
Therefore, using Pythagorean Theorem to find 𝑑:
2
2
2
2
3.) An air balloon situated 150m horizontally from an observer fly vertically upward with
constant velocity. After 1min, its angle of elevation 28°59’. What will be its angle of elevation
after 3min from the initial position?
Solving for 𝛼:
tan 𝛼 =
3 [ 150 tan(28°59′)]
𝛼 = tan
− 1
[
150 tan
)]
Thus, the angle of elevation after 3 minutes is approximately 58.9621°.
4.) From the top of lighthouse, 53m above the water, the angle of depression of a boat due south
is 18°50’. Calculate the speed of the boat if after it moves due west for 2 minutes the angle of
depression is 14° 20 ’.
Let 𝛼 be the angle of depression from the top of lighthouse to the boat due south given as 18 °50’.
Let 𝛽 be the angle of depression from the top of lighthouse to the boat due west given as 14 ° 2 0’.
Let 𝑡 be the time given as 2 minutes.
Let 𝑥 be the distance traveled of boat due west for 2 minutes.
Let 𝑠 be the speed of the boat.
Solution:
Referring from △ 𝐵𝐷𝐶:
tan 𝛼 = tan(18°50’) =
→ 𝑦 = 53 tan( 18 ° 50 ’)
Referring from △ 𝐴𝐷𝐶:
tan 𝛽 = tan
→ 𝑧 = 53 tan
Assuming that the boat travels at a constant velocity and referring from △ 𝐴𝐶𝐵 it is a right
triangle. Thus, Using Pythagorean Theorem to solve for 𝑥:
2
2
= √( 53 [tan(14°20’)])
2
− ( 53 [tan(18°50’)])
2
Assuming that the boat travels at a constant velocity,
[
tan
)])
2
[
tan
)])
2
Or
√( 53 [tan( 14 ° 20 ’)])
2
− ( 53 [tan( 18 ° 50 ’)])
2
×
5.) A flagpole on the edge of the bank of a river. From a point on the opposite bank directly across
from the flagpole, the measure of the angle of elevation to the top of the pole is 25°. From the point
200 ft further away and in line with the pole and the first point, the measure of the angle of elevation
to the top of the pole is 21°. Find the distance across the river in ft.
Let 𝛼 be the angle equal to 25°.
Let 𝛽 be the angle equal to 21°.
Let 𝑥 be the distance across the river in ft.
Let ℎ be the height of the flagpole.
Solution:
