Exercise (11.3) - No(1) Sample Solutions
(b) Solution (1)
sin135∘=sin(180∘−45∘)=sin45∘=√22
cos135∘ =cos(180∘−45∘)=−cos45∘=−√22
tan135∘ =−1cot135∘ =−1sec135∘ =−√2cosec135∘ =√2
(b) Solution (2)
principal angle = 135∘∴basic acute angle = 45∘
∴ sin45∘=√22and cos45∘=√22
sin135∘=sin45∘and cos135∘=cos45∘ numerically. But 135∘ lies in the second quadrant.
∴sin135∘=sin45∘=√22
cos135∘=−cos45∘=−√22
tan135∘ =−1cot135∘ =−1sec135∘ =−√2cosec135∘ =√2
(d) Solution (1)
sin210∘=sin(180∘+30∘)=−sin30∘=−12
cos210∘=cos(180∘+30∘)=−cos30∘=−√32
tan210∘=√33
cot210∘=√3
sec210∘=−2√33
cosec210∘=−2
(d) Solution (2)
principal angle = 210∘
∴basic acute angle = 30∘
∴ sin30∘=12and cos30∘=√32
sin210∘=sin30∘and cos210∘=cos30∘ numerically.
But 210∘ lies in the third quadrant.
∴sin210∘=sin30∘=−12
cos210∘=−cos30∘=−√32
tan210∘ =√33cot210∘ =√3sec210∘ =−2√33cosec210∘ =−2
(m) Solution (1)
sin(−120∘)=−sin120∘
=−sin(180∘−60∘)
=−sin60∘
=−√32
cos(−120∘)=cos120∘
=cos(180∘−60∘)
=−cos60∘
=−12
tan(−120∘)=√3
cot(−120∘)=√33
sec(−120∘)=−2
cosec(−120∘)=−2√33
(m) Solution (2)
principal angle = −120∘
∴basic acute angle = 60∘
∴ sin60∘=√32and cos60∘=12
sin(−120∘)=sin60∘and cos(−120∘)x=cos60∘ numerically.
But (−120∘) lies in the third quadrant.
∴sin(−120∘)=sin60∘=−√32
cos(−120∘)=−cos60∘=−12
tan(−120∘)=√3
cot(−120∘)=√33
sec(−120∘)=−2
cosec(−120∘)=−2√33
(q) Solution (1)
sin480∘=sin(360∘+60∘)
=sin60∘
=√32
cos480∘=cos(360∘+60∘)
=cos60∘
=12
tan480∘=√3
cot480∘=√33
sec480∘=2
cosec480∘=2√33
(q) Solution (2)
principal angle = 480∘
∴basic acute angle = 60∘
∴ sin60∘=√32and cos60∘=12
sin480∘=sin60∘and cos480∘=cos60∘ numerically.
But 480∘ lies in the first quadrant.
∴sin480∘=sin60∘=√32
cos480∘=cos30∘=12
tan480∘=√3
cot480∘=√33
sec480∘=2
cosec480∘=2√33
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