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三角比ver2

Authored by 田原恭比古 田原恭比古

Mathematics

10th Grade

Used 3+ times

三角比ver2
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11 questions

Show all answers

1.

MULTIPLE SELECT QUESTION

45 sec • 5 pts

12\frac{1}{\sqrt[]{2}}  となる三角比をすべて選べ。

sin45°

cos45°

sin135°

cos135°

2.

MULTIPLE SELECT QUESTION

45 sec • 5 pts

   cos⁡θ=−32\cos\theta=-\frac{\sqrt{3}}{2}  となる値を 0∘≤θ≤180∘0^{\circ}\le\theta\le180^{\circ}  で求めよ。

θ=30∘\theta=30^{\circ}  

θ=150∘\theta=150^{\circ}

θ=60∘\theta=60^{\circ}

θ=120∘\theta=120^{\circ}

3.

MULTIPLE CHOICE QUESTION

30 sec • 5 pts

3tan⁡θ−1=0\sqrt{3}\tan\theta-1=0  を満たす θ\theta  を 0∘≤θ≤180∘0^{\circ}\le\theta\le180^{\circ} で求めよ。

θ=30∘\theta=30^{\circ}

θ=60∘\theta=60^{\circ}

θ=120∘\theta=120^{\circ}

θ=135∘\theta=135^{\circ}

4.

MULTIPLE CHOICE QUESTION

30 sec • 5 pts

0∘<θ<180∘0^{\circ}<\theta<180^{\circ}  とする。 sin⁡θcos⁡θ<0\sin\theta\cos\theta<0  を満たすとき, tan⁡θ\tan\theta  の符号は?

正

負

0

5.

MULTIPLE SELECT QUESTION

45 sec • 5 pts

三角比の相互関係をすべて選べ。

sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1  

tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}  

tan⁡2θ=1−cos⁡2θcos⁡2θ\tan^2\theta=\frac{1-\cos^2\theta}{\cos^2\theta}  

1−tan⁡2θ=1cos⁡2θ1-\tan^2\theta=\frac{1}{\cos^2\theta}  

6.

MULTIPLE CHOICE QUESTION

45 sec • 5 pts

0∘≤θ≤180∘0^{\circ}\le\theta\le180^{\circ}   とする。 cos⁡θ=45\cos\theta=\frac{4}{5}   を満たすとき, sin⁡θ\sin\theta   の値は?

sin⁡θ=14\sin\theta=\frac{1}{4}  

sin⁡θ=43\sin\theta=\frac{4}{3}

sin⁡θ=35\sin\theta=\frac{3}{5}

sin⁡θ=34\sin\theta=\frac{3}{4}

7.

MULTIPLE CHOICE QUESTION

30 sec • 5 pts

tan⁡θ=−2\tan\theta=-2  のとき、 cos⁡θ\cos\theta  の値は?

cos⁡θ=15\cos\theta=\frac{1}{\sqrt{5}}  

cos⁡θ=25\cos\theta=\frac{2}{\sqrt{5}}

cos⁡θ=−15\cos\theta=-\frac{1}{\sqrt{5}}

cos⁡θ=13\cos\theta=\frac{1}{\sqrt{3}}

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