Find the value of θ, 0° ≤ θ ≤ 360° for the following equations. Do not use table.
(a) sinθ=−12
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(b) cosθ=−√32
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(c) cosθ=−1√2
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(d) tanθ=√3
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(e) tan2θ=1
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(f) tan3θ=−1
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∴ basic acute angle=45∘
But tan3θ is negative, 3θ may be either in the 2nd or 4th quadrant.
For 2nd quadrant_,
3θ=180∘−45∘
3θ=135∘
θ=45∘
(or)
3θ=360∘+180∘−45∘
3θ=495∘
θ=165∘
(or)
3θ=360∘+360∘+180∘−45∘
3θ=855∘
θ=285∘
For 4th quadrant_,
3θ=360∘−45∘
3θ=315∘
θ=105∘
(or)
3θ=360∘+360∘−45∘
3θ=675∘
θ=225∘
(or)
3θ=360∘+360∘+360∘−45∘
3θ=1035∘
θ=345∘
But tan3θ is negative, 3θ may be either in the 2nd or 4th quadrant.
For 2nd quadrant_,
3θ=180∘−45∘
3θ=135∘
θ=45∘
(or)
3θ=360∘+180∘−45∘
3θ=495∘
θ=165∘
(or)
3θ=360∘+360∘+180∘−45∘
3θ=855∘
θ=285∘
For 4th quadrant_,
3θ=360∘−45∘
3θ=315∘
θ=105∘
(or)
3θ=360∘+360∘−45∘
3θ=675∘
θ=225∘
(or)
3θ=360∘+360∘+360∘−45∘
3θ=1035∘
θ=345∘
diagram မ်ားသည္ principal angle မ်ား ကို ဆံုးျဖတ္ရာတြင္ အေထာက္အကူ ျဖစ္ေစရန္ ေရးဆြဲ ေဖၚျပျခင္း ျဖစ္သည္။ နားလည္ ကၽြမ္းက်င္ သြားလ်င္ diagram မ်ား မေရးဆြဲပဲ ေျဖဆိုႏိုင္ ပါသည္။ က်န္ေသာပုစာၦမ်ားအတြက္ diagram မဆြဲေတာ့ပဲ ေျဖဆိုပါမည္။
(g) tan(3θ−30∘)=−1
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∴ basic acute angle=45∘ But tan(3θ−30∘) is negative, (3θ−30∘) may be either in the 2nd or 4th quadrant.For 2nd quadrant_, 3θ−30∘=180∘−45∘ 3θ=165∘ θ=55∘ (or) 3θ−30∘=360∘+180∘−45∘ 3θ=525∘ θ=175∘ (or) 3θ−30∘=360∘+360∘+180∘−45∘ 3θ=885∘ θ=295∘ For 4th quadrant_, 3θ−30∘=360∘−45∘ 3θ=345∘ θ=115∘ (or) 3θ−30∘=360∘+360∘−45∘ 3θ=705∘ θ=295∘ (or) 3θ−30∘=360∘+360∘+360∘−45∘ 3θ=1065∘ θ=355∘
(h) cos2θ=−12
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∴ basic acute angle=60∘ But cos2θ is negative, 2θ may be either in the 2nd or 3rd quadrant.For 2nd quadrant_, 2θ=180∘−60∘ 2θ=120∘ θ=60∘ (or) 2θ=360∘+180∘−60∘ 2θ=480∘ θ=240∘ For 3rd quadrant_, 2θ=180∘+60∘ 2θ=240∘ θ=120∘ (or) 2θ=360∘+180∘+60∘ 2θ=600∘ θ=300∘
(i) sin(2θ+30∘)=√32
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∴ basic acute angle=60∘ But sin(2θ+30∘) is positive, 2θ+30∘ may be either in the 1st or 2nd quadrant.For 1st quadrant_, 2θ+30∘=60∘ 2θ=30∘ θ=15∘ (or) 2θ=360∘+60∘ 2θ=420∘ θ=210∘ For 2nd quadrant_, 2θ+30∘=180∘−60∘ 2θ=120∘ θ=60∘ (or) 2θ=360∘+180∘−60∘ 2θ=480∘ θ=240∘
(j) tan12θ=−1√3
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∴ basic acute angle=30∘
But tan12θ is positive, 12θ may be
either in the 2nd or 4th quadrant.
∴ 12θ=180∘−30∘ (or) 12θ=360∘−30∘
∴ 12θ=150∘ (or) 12θ=330∘
∴ θ=300∘ (or) θ=660∘
But 0∘≤θ≤360∘,θ=660∘ is not in domain.∴ θ=300∘ is the only solution.
But tan12θ is positive, 12θ may be
either in the 2nd or 4th quadrant.
∴ 12θ=180∘−30∘ (or) 12θ=360∘−30∘
∴ 12θ=150∘ (or) 12θ=330∘
∴ θ=300∘ (or) θ=660∘
But 0∘≤θ≤360∘,θ=660∘ is not in domain.∴ θ=300∘ is the only solution.
(k) sinθ=0.6521
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∴ basic acute angle=40∘42′ (using table) But sinθ is positive, θ may be either in the 1st or 2nd quadrant.∴ θ=40∘42′ (or) θ=180∘−40∘42′∴ θ=40∘42′ (or) θ=139∘18′
(l) cosθ=−0.3854
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∴ basic acute angle=67∘20′ (using table) But cosθ is negative, θ may be either in the 2nd or 3rd quadrant.∴ θ=180∘−67∘20′ (or) θ=180∘+67∘20′∴ θ=112∘40′ (or) θ=247∘20′
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