问题1.评估：
(i)正弦18°/余弦72°
解决方案：
Since,
cos 72° = cos ( 90° – 18° ) = sin 18°
Therefore,
sin 18° / cos 72° = sin 18° / sin 18° = 1
Hence, sin 18° / cos 72° = 1.
(ii)棕褐色26°/婴儿床64°
解决方案：
Since,
cot 64° = cot ( 90° – 26° ) = tan 26°
Therefore,
tan 26° / cot 64° = tan 26° / tan 26° = 1
Hence, tan 26° / cot 64° = 1.
(iii)cos 48°–正弦42°
Since,
cos 48° = cos ( 90° – 42° ) = sin 42°
Therefore,
cos 48° – sin 42° = sin 42° – sin 42° = 0
Hence, cos 48° – sin 42° = 0.
(iv)cosec 31°– sec 59°
解决方案：
Since,
sec 59° = sec ( 90° – 31° ) = cosec 31°
Therefore ,
cosec 31° – sec 59° = cosec 31° – cosec 31° = 0
Hence, cosec 31° – sec 59° = 0.
问题2：表明：
(i)棕褐色48°棕褐色23°棕褐色42°棕褐色67°= 1
解决方案：
Let A = tan 48° tan 23° tan 42° tan 67°
Since ,
tan 23° = tan( 90° – 23° ) = cot 67° and,
tan 42° = cot( 90° – 42° ) = cot 48°
Therefore,
A = tan 48° cot 67° cot 48° tan 67°
A = 1 (Since, tan B° cot B° = 1)
Hence, tan 48° tan 23° tan 42° tan 67° = 1
(ii)cos 38°cos 52°–正弦38°正弦52°= 0
Let A = cos 38° cos 52° – sin 38° sin 52°
Since,
sin 52° = sin (90° – 38°) = cos 38° and,
cos 52° = cos(90° – 52°) = sin 38°
Therefore,
A = cos 38° sin 38° – sin 38° cos 38°
A = 0
Hence, cos 38° cos 52° – sin 38° sin 52° = 0.
问题3.如果tan 2A = cot(A – 18°)，其中2A是锐角，则求A的值。
解决方案：
We have,
tan 2A = cot ( A – 18° ) —(1)
Since,
tan (2A) = cot ( 90° – 2A ) — (2)
Putting (2) in (1),
cot ( 90° – 2A ) = cot ( A – 18° )
Therefore,
90° – 2A = A – 18°
3A = 108°
A = 36°
Hence, A = 36°.
问题4.如果tan A =婴儿床B，则证明A + B = 90°。
解决方案：
We have,
tan A = cot B —(1)
Since,
tan (A) = cot (90° – A) — (2)
Putting (2) in (1),
cot (90° – A) = cot (B)
Therefore,
90° – A = B
Hence, A + B = 90°.
问题5.如果sec 4A = sec(A – 20°)，其中4A是锐角，则求A的值。
解决方案：
We have,
sec 4A = cosec ( A – 20° ) —(1)
Since,
sec 4A = cosec ( 90° – 4A ) — (2)
Putting (2) in (1),
cosec ( 90° – 4A ) = cosec ( A – 20° )
Therefore,
90° – 4A = A – 20°
5A = 110°
A = 22°
Hence, A = 22°.
问题6.如果A，B和C是三角形ABC的内角，则表明sin((B + C)/ 2)= cos(A / 2)。
解决方案：
Let T = sin ((B + C) / 2) — (1)
A, B and C are the interior angles of triangle ABC, therefore,
A + B + C = 180°
Dividing by 2 on both sides
(B + C)/2 = 90° – (A / 2) —(2)
Putting (2) on (1)
T = sin (90° – (A / 2)
= cos (A / 2)
Hence, sin ((B + C)/2) = cos (A / 2).
问题7.以0°至45°的角度的三角比表示sin 67°+ cos 75°
解决方案：
Let A = sin 67° + cos 75°
Since,
sin 67° = sin(90° – 23°) = cos (23°)
cos 75° = cos (90° – 15°) = sin (15°)
Therefore,
sin 67° + cos 75° = cos 23° + sin 15°