Nilai dari ( tan 1 0 ∘ + tan 7 0 ∘ − tan 5 0 ∘ ) adalah ....

Pembahasan

Ingat rumus jumlah dan selisihtrigonometri berikut ini: 2 sin A cos B = sin ( A + B ) + sin ( A − B ) 2 cos A sin B = sin ( A + B ) − sin ( A − B ) 2 cos A cos B = cos ( A + B ) + cos ( A − B ) sin A − sin B = 2 cos 2 1 ​ ( A + B ) sin 2 1 ​ ( A − B ) cos A + cos B = 2 cos 2 1 ​ ( A + B ) cos 2 1 ​ ( A − B ) Dengan menggunakan konsep di atas, diperoleh hasil: tan 1 0 ∘ + tan 7 0 ∘ − tan 5 0 ∘ = ( c o s 1 0 ∘ s i n 1 0 ∘ ​ + c o s 7 0 ∘ s i n 7 0 ∘ ​ ) − c o s 5 0 ∘ s i n 5 0 ∘ ​ = ( c o s 7 0 ∘ c o s 1 0 ∘ c o s 7 0 ∘ s i n 1 0 ∘ + s i n 7 0 ∘ c o s 1 0 ∘ ​ ) − c o s 5 0 ∘ s i n 5 0 ∘ ​ = 2 2 ​ ⋅ ( c o s 7 0 ∘ c o s 1 0 ∘ c o s 7 0 ∘ s i n 1 0 ∘ + s i n 7 0 ∘ c o s 1 0 ∘ ​ ) − c o s 5 0 ∘ s i n 5 0 ∘ ​ = ( 2 c o s 7 0 ∘ c o s 1 0 ∘ 2 c o s 7 0 ∘ s i n 1 0 ∘ + 2 s i n 7 0 ∘ c o s 1 0 ∘ ​ ) − c o s 5 0 ∘ s i n 5 0 ∘ ​ = ( c o s ( 7 0 ∘ + 1 0 ∘ ) + c o s ( 7 0 ∘ − 1 0 ∘ ) ( s i n ( 7 0 ∘ + 1 0 ∘ ) − s i n ( 7 0 ∘ − 1 0 ∘ ) ) + ( s i n ( 7 0 ∘ + 1 0 ∘ ) + s i n ( 7 0 ∘ − 1 0 ∘ ) ) ​ ) − c o s 5 0 ∘ s i n 5 0 ∘ ​ = ( c o s 8 0 ∘ + c o s 6 0 ∘ ( s i n 8 0 ∘ − s i n 6 0 ∘ ) + ( s i n 8 0 ∘ + s i n 6 0 ∘ ) ​ ) − c o s 5 0 ∘ s i n 5 0 ∘ ​ = ⎝ ⎛ ​ c o s 8 0 ∘ + 2 1 ​ 2 s i n 8 0 ∘ ​ ⎠ ⎞ ​ − c o s 5 0 ∘ s i n 5 0 ∘ ​ = c o s 8 0 ∘ c o s 5 0 ∘ + 2 1 ​ cos 5 0 ∘ 2 s i n 8 0 ∘ c o s 5 0 ∘ − s i n 5 0 ∘ ( c o s 8 0 ∘ + 2 1 ​ ) ​ ( samakan penyebut ) = c o s 8 0 ∘ c o s 5 0 ∘ + 2 1 ​ cos 5 0 ∘ 2 s i n 8 0 ∘ c o s 5 0 ∘ − c o s 8 0 ∘ s i n 5 0 ∘ − 2 1 ​ s i n 5 0 ∘ ​ = c o s 8 0 ∘ c o s 5 0 ∘ + 2 1 ​ cos 5 0 ∘ 2 s i n 8 0 ∘ c o s 5 0 ∘ − c o s 8 0 ∘ s i n 5 0 ∘ − 2 1 ​ s i n 5 0 ∘ ​ ⋅ 2 2 ​ = 2 c o s 8 0 ∘ c o s 5 0 ∘ + cos 5 0 ∘ 2 ⋅ 2 s i n 8 0 ∘ c o s 5 0 ∘ − 2 c o s 8 0 ∘ s i n 5 0 ∘ − s i n 5 0 ∘ ​ = c o s ( 8 0 ∘ + 5 0 ∘ ) + c o s ( 8 0 ∘ − 5 0 ∘ ) + cos 5 0 ∘ 2 ( s i n ( 8 0 ∘ + 5 0 ∘ ) + s i n ( 8 0 ∘ − 5 0 ∘ ) ) − ( s i n ( 8 0 ∘ + 5 0 ∘ ) − s i n ( 8 0 ∘ − 5 0 ∘ ) ) − s i n 5 0 ∘ ​ = c o s 13 0 ∘ + c o s 3 0 ∘ + cos 5 0 ∘ 2 ( s i n 13 0 ∘ + s i n 3 0 ∘ ) − ( s i n 13 0 ∘ − s i n 3 0 ∘ ) − s i n 5 0 ∘ ​ = c o s 13 0 ∘ + 2 1 ​ 3 ​ + cos 5 0 ∘ 2 ( s i n 13 0 ∘ + 2 1 ​ ) − ( s i n 13 0 ∘ − 2 1 ​ ) − s i n 5 0 ∘ ​ = 2 1 ​ 3 ​ + c o s 13 0 ∘ + cos 5 0 ∘ 2 s i n 13 0 ∘ + 1 − s i n 13 0 ∘ + 2 1 ​ − s i n 5 0 ∘ ​ = 2 1 ​ 3 ​ + c o s 13 0 ∘ + cos 5 0 ∘ 2 3 ​ + s i n 13 0 ∘ − s i n 5 0 ∘ ​ = 2 1 ​ 3 ​ + ( 2 c o s 2 1 ​ ( 13 0 ∘ + 5 0 ∘ ) c o s 2 1 ​ ( 13 0 ∘ − 5 0 ∘ ) ) 2 3 ​ + ( 2 c o s 2 1 ​ ( 13 0 ∘ + 5 0 ∘ ) s i n 2 1 ​ ( 13 0 ∘ − 5 0 ∘ ) ) ​ = 2 1 ​ 3 ​ + ( 2 c o s 2 1 ​ ( 18 0 ∘ ) c o s 2 1 ​ ( 8 0 ∘ ) ) 2 3 ​ + ( 2 c o s 2 1 ​ ( 18 0 ∘ ) s i n 2 1 ​ ( 8 0 ∘ ) ) ​ = 2 1 ​ 3 ​ + ( 2 c o s 9 0 ∘ c o s 4 0 ∘ ) 2 3 ​ + ( 2 c o s 9 0 ∘ s i n 4 0 ∘ ) ​ = 2 1 ​ 3 ​ + ( 2 ⋅ 0 ⋅ c o s 4 0 ∘ ) 2 3 ​ + ( 2 ⋅ 0 ⋅ s i n 4 0 ∘ ) ​ = 2 1 ​ 3 ​ + 0 2 3 ​ + 0 ​ = 2 1 ​ 3 ​ 2 3 ​ ​ = 2 3 ​ ÷ 2 3 ​ ​ = 2 3 ​ × 3 ​ 2 ​ = 3 ​ 3 ​ = 3 ​ 3 ​ × 3 ​ 3 ​ ​ = 3 3 3 ​ ​ = 3 ​ ​ Jadi, tan 1 0 ∘ + tan 7 0 ∘ − tan 5 0 ∘ = ​ ​ 3 ​ ​ . Oleh karena itu, jawaban yang benar adalah D.