Homework PracticeSign Up
Back to assignmentPrintable assignment view
Print options

Homework Practice

Printable assignment

Introduction to Trigonometry Practice 2: Standard Values and Relationships

Section A: Standard and Boundary Angles

Question 1mcq
What is the exact value of sin30\sin 30^\circ?
A.12\frac{1}{\sqrt{2}}
B.11
C.12\frac{1}{2}
D.32\frac{\sqrt{3}}{2}
Question 2mcq
Which pair of values is correct?
A.tan60=1\tan 60^\circ=1 and cos45=12\cos 45^\circ=\frac{1}{2}
B.cos60=12\cos 60^\circ=\frac{1}{2} and tan45=1\tan 45^\circ=1
C.sin60=12\sin 60^\circ=\frac{1}{2} and tan45=0\tan 45^\circ=0
D.cos30=12\cos 30^\circ=\frac{1}{2} and sin45=1\sin 45^\circ=1
Question 3mcq
Which value is defined and correct?
A.cos0=1\cos 0^\circ=1
B.tan90=0\tan 90^\circ=0
C.sec90=1\sec 90^\circ=1
D.cosec0=1\cosec 0^\circ=1
Question 4mcq
Evaluate 2sin30+cos602\sin 30^\circ+\cos 60^\circ.
A.11
B.22
C.12\frac{1}{2}
D.32\frac{3}{2}
Question 5mcq
Evaluate tan245+cos260\tan^2 45^\circ+\cos^2 60^\circ.
A.11
B.14\frac{1}{4}
C.54\frac{5}{4}
D.34\frac{3}{4}

Section B: Relationships

Question 6mcq
If A=35A=35^\circ, then sin(90A)\sin(90^\circ-A) equals:
A.sec35\sec 35^\circ
B.cos35\cos 35^\circ
C.sin35\sin 35^\circ
D.tan35\tan 35^\circ
Question 7mcq
Which identity is correct for acute angle AA?
A.cot(90A)=tanA\cot(90^\circ-A)=\tan A
B.sin(90A)=sinA\sin(90^\circ-A)=\sin A
C.sec(90A)=secA\sec(90^\circ-A)=\sec A
D.tan(90A)=tanA\tan(90^\circ-A)=\tan A
Question 8mcq
If secA=53\sec A=\frac{5}{3}, what is cosA\cos A?
A.53\frac{5}{3}
B.45\frac{4}{5}
C.54\frac{5}{4}
D.35\frac{3}{5}
Question 9mcq
Which reciprocal relationship is correct?
A.cosecA=1cosA\cosec A=\frac{1}{\cos A}
B.tanA=1sinA\tan A=\frac{1}{\sin A}
C.cotA=1tanA\cot A=\frac{1}{\tan A}
D.secA=1sinA\sec A=\frac{1}{\sin A}
Question 10mcq
A student writes tan90=sin90cos90=10\tan 90^\circ=\frac{\sin 90^\circ}{\cos 90^\circ}=\frac{1}{0} and calls it 00. What should be concluded?
A.tan90=1\tan 90^\circ=-1
B.tan90\tan 90^\circ is not defined.
C.tan90=0\tan 90^\circ=0
D.tan90=1\tan 90^\circ=1