sin 15° – cos 15° ———————— = Tan 15°
1. sin 15° – cos 15° ———————— = Tan 15°
silakan dicermati ya
2. sin 15 - cos 15 per tan 15 =
0,995005147099 semoga membantu
3. sin 15 - cos 15 : Tan 15
Jawab:
Penjelasan dengan langkah-langkah:
tan = sin : cos
sin^2 a + cos^2 a = 1
sin 15 - (cos 15 : tan 15)
= sin 15 - (cos 15 : (sin 15 : cos 15))
= sin 15 - (cos 15 x (cos 15 : sin 15))
= sin 15 - (cos^2 15 : sin 15)
disamakan penyebutnya,
= sin^2 15 - cos^2 15
= -1
4. sin cos tan dari 15 derajat
Bab Trigonometri
Matematika SMA Kelas X
1] sin 15
= sin (60 - 45)
= sin 60 cos 45 - cos 60 sin 45
= 1/2 akar 3 . 1/2 akar 2 - 1/2 . 1/2 akar 2
= 1/4 akar 2 (akar 3 - 1)
2] cos 15
= cos (60 - 45)
= cos 60 cos 45 + sin 60 sin 45
= 1/2 . 1/2 akar 2 + 1/2 akar 3 . 1/2 akar 2
= 1/4 akar 2 (1 + akar 3)
= 1/4 akar 2 (akar 3 + 1)
3] tan 15
= (sin 15) : (cos 15)
= (1/4 akar 2 (akar 3 - 1)) : (1/4 akar 2(akar 3 + 1))
= (akar 3 - 1)\(akar 3 + 1)
= (3 - 2 akar 3 + 1)/(3 - 1)
= 2 - akar 3
5. nilai Sin 15 cos 15 tan 15 sm sin 75 cos 75 dan tan 75?
sin15: 0,2588 cos15: 0,9659 tan15: 0,2679 sin75: 0,9659 cos75: 0,2588 tan75: 3,7320
6. Sin 105° + tan 15° ————————- = ——— adalah... tan 75° - cos 15°
mari ki maaf kalau salah
7. Nilai trigonometri 4 sin 15 sin 22,5 cos 22,5 per cos 15 (1-tan² 15)
Jawaban:
hasilnya
[tex] \frac{ \sqrt{6} }{6} [/tex]
maaf kalo salah yaa hehe
8. tan 15° Sin 75° Cos 15
Jawab:
¼
Penjelasan dengan langkah-langkah:
[tex]\displaystyle \tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}\\\tan 15^\circ=\tan (45^\circ-30^\circ)\\=\frac{\tan 45^\circ-\tan 30^\circ}{1+\tan 45^\circ \tan 30^\circ}\\=\frac{1-\frac{\sqrt{3}}{3}}{1+1\left ( \frac{\sqrt{3}}{3} \right )}\\=\frac{3-\sqrt{3}}{3+\sqrt{3}}\\=\frac{3-\sqrt{3}}{3+\sqrt{3}}~\frac{3-\sqrt{3}}{3-\sqrt{3}}\\=\frac{12-6\sqrt{3}}{6}\\=2-\sqrt{3}[/tex]
[tex]\displaystyle 2\sin \alpha \cos \beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)\\\sin 75^\circ \cos 15^\circ=\frac{\sin 90^\circ+\sin 60^\circ}{2}\\=\frac{1+\frac{\sqrt{3}}{2}}{2}\\=\frac{2+\sqrt{3}}{4}[/tex]
[tex]\displaystyle \tan 15^\circ \sin 75^\circ \cos 15^\circ\\=\left ( 2-\sqrt{3} \right )\left ( \frac{2+\sqrt{3}}{4} \right )\\=\frac{4-3}{4}\\=\frac{1}{4}=0,25=2^{-2}[/tex]
9. sin 15 ,cos 15 ,tan 15 derajat
sin 15 = 0,258 = 0,26
cos 15 = 0,965 = 0,97
tan 15 = 0,267 = 0,27
Itu terserah kamu mau ambil 2 angka pertama setelah koma tanpa pembulatan atau yang telah dibulatkan^^
sin 15 = 0,650
cos 15 = -0,759
tan 15 = -0,855
10. Nilai dari sin 15° - cos 15° /(per) tan adalah
Jawaban:
nilai dari sin 15°-cos 15°/(per) tan adalah?
Penjelasan dengan langkah-langkah:
semoga bermanfaat
11. nilai cos 255° sama dengan a. - sin 15°b. - cos 15c. - tan 15° d. sin 15°e. cos 15°
Penjelasan dengan langkah-langkah:
sin 255° = - cos (270 - 255)°
sin 255° = - cos 15°
Semoga Bermanfaat
12. sin 15⁰- cos 15⁰ / tan 15⁰
Jawab:
6353
Penjelasan dengan langkah-langkah:
a
13. sin 15- cos 15/ tan 15
sin 15 - cos 15 / tan 15
sin 15 - cos 15 / sin 15 /cos 15
sin 15 - (cos 15 x cos 15/sin 15)
sin 15 - cos²15/sin 15
sin²15 / sin 15 - cos²15 / sin 15
sin²15-1-sin²15/sin 15
= -1-2sin²15/sin 15
14. cara menentukan sin 15°, cos 15° dan tan 15°
sin 15° adalah [tex]\frac{1}{4}(\sqrt{6}-\sqrt{2})}[/tex]
cos 15° adalah [tex]\frac{1}{4}(\sqrt{2}+\sqrt{6})}[/tex]
tan 15° adalah [tex]2-\sqrt{3}}[/tex]
PEMBAHASAN
Untuk mengerjakan soal di atas kita dapat menggunakan rumus trigonometri selisih 2 sudut.
[tex]\boxed{\bold{sin~(\alpha-\beta)=sin~\alpha.cos~\beta-cos~\alpha.sin~\beta}}\\\\\boxed{\bold{cos~(\alpha-\beta)=cos~\alpha.cos~\beta+sin~\alpha.sin~\beta}}\\\\\boxed{\bold{tan~(\alpha-\beta)=\frac{tan~\alpha-tan~\beta}{1+tan~\alpha.tan~\beta}}}\\[/tex]
==============================================================
Pertama, kita ubah dulu 15° ke dalam sudut istimewa!
15° = 60° - 45°
sin 15° = sin (60° - 45°)
cos 15° = sin (60° - 45°)
tan 15° = sin (60° - 45°)
sin 15°
[tex]sin~15^o=sin~(60^o-45^o)\\sin~15^o=sin~60^o.cos~45^o-cos~60^o.sin~45^o\\\\sin~15^o=\frac{1}{2} \sqrt{3}.\frac{1}{2} \sqrt{2}-\frac{1}{2}.\frac{1}{2} \sqrt{2}\\ \\sin~15^o=\frac{1}{4} \sqrt{6}-\frac{1}{4} \sqrt{2}\\\\\boxed{sin~15^o=\frac{1}{4}(\sqrt{6}-\sqrt{2})}[/tex]
cos 15°
[tex]cos~15^o=cos~(60^o-45^o)\\\\cos~15^o=cos~60^o.cos~45^o+sin~60^o.sin~45^o\\\\cos~15^o=\frac{1}{2}.\frac{1}{2} \sqrt{2}+\frac{1}{2} \sqrt{3}.\frac{1}{2} \sqrt{2}\\\\cos~15^o=\frac{1}{4} \sqrt{2}+\frac{1}{4} \sqrt{6}\\\\\boxed{cos~15^o=\frac{1}{4}(\sqrt{2}+\sqrt{6})}[/tex]
tan 15°
[tex]tan~(15^o)=tan~(60^o-45^o)\\tan~(15^o)=\frac{tan~60^o-tan~45^o}{1+tan~60^o.tan~45^o}\\\\tan~(15^o)=\frac{\sqrt{3}-1 }{1+\sqrt{3}.1}\\\\tan~(15^o)=\frac{\sqrt{3}-1 }{\sqrt{3}+1}[/tex]
Rasionalkan bentuk akar di atas menjadi :
[tex]tan~(15^o)=\frac{\sqrt{3}-1 }{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}\\ \\tan~(15^o)=\frac{3-2\sqrt{3} +1}{3-1}\\ \\tan~(15^o)=\frac{4-2\sqrt{3}}{2}\\\\tan~(15^o)=\frac{2(2-\sqrt{3})}{2}\\\\\boxed{tan~(15^o)=2-\sqrt{3}}[/tex]
Pelajari soal serupa :
brainly.co.id/tugas/1685827
brainly.co.id/tugas/6329482
Detail Jawaban
Kelas : 10
Mapel : Matematika
Materi : Trigonometri
Kode kategorisasi : 10.2.7
Kata kunci : sinus, cosinus, tangen, selisih sudut, trigonometri.
#optitimcompetition
15. sin 15° - cos 15° / tan 15°
Penjelasan dengan langkah-langkah:
sin 15° - cos 15°/tan15°
~> sin 15° = sin (45° - 30°)
= sin45.cos30 - sin30.cos45
= √2/2.√3/2 - 1/2.√2/2
= √6/4 - √2/4 ~ (√6 - √2)/4 ✓
~> cos 15° = cos (45° - 30°)
= cos45.cos30 + sin45.sin30
= √2/2.√3/2 + √2/2.1/2
= √6/4 + √2/4 ~ (√6 + √2)/4 ✓
~> tan 15° = tan (45° - 30°)
= (tan45 - tan30)/1 + tan45.tan30
= (1 - 1/√3)/1 + 1/√3
= (√3 - 1)/√3/(√3 + 1)/√3
= (√3 - 1)/(√3 + 1) ✓
~> [(√6 - √2)/4 - (√6 + √2)/4]/[(√3 - 1)/(√3 + 1)]
-(√3 + 1)√2/2(√3 - 1)
(√6 + √2)/(2 - 2√3) ✓✓
16. Nilai dari sin (15° - cos 15°) : Tan 15° adalah....
Jawaban:
sin 15° - cos 15°/tan15°
~> sin 15° = sin (45° - 30°)
= sin45.cos30 - sin30.cos45
= √2/2.√3/2 - 1/2.√2/2
= √6/4 - √2/4 ~ (√6 - √2)/4 ✓
~> cos 15° = cos (45° - 30°)
= cos45.cos30 + sin45.sin30
= √2/2.√3/2 + √2/2.1/2
= √6/4 + √2/4 ~ (√6 + √2)/4 ✓
~> tan 15° = tan (45° - 30°)
= (tan45 - tan30)/1 + tan45.tan30
= (1 - 1/√3)/1 + 1/√3
= (√3 - 1)/√3/(√3 + 1)/√3
= (√3 - 1)/(√3 + 1) ✓
~> [(√6 - √2)/4 - (√6 + √2)/4]/[(√3 - 1)/(√3 + 1)]
-(√3 + 1)√2/2(√3 - 1)
(√6 + √2)/(2 - 2√3) ✓✓
maaf kalau salah
17. nilai sin 15° + cos 15° / tan 240° adalah....
Jawabannya -0,03
Maaf ya...klo salah....( sin 15° + cos 15° ) / tan 240°
untuk sin 15° = sin (45 - 30)°
= sin 45 cos 30 - cos 45 sin 30
= (½√2) (½√3) - (½√2) (½)
= ¼ √6 - (¼ √2)
= ¼ (√6 - √2)
untuk cos 15° = cos (45 - 30)°
= cos 45° cos 30° + sin 45° sin 30°
= (½√2) (½√3) + (½√2) (½)
= ¼ (√6) + ¼√2
= ¼ (√6 + √2)
untuk tan 240° = √3
sin 15° + cos 15°
= ¼ (√6 - √2) + (¼ (√6 + √2)
= ½√6
maka nilai sin 15° + cos 15° / tan 240°
[tex] \frac{ \frac{1}{2} \sqrt{6} }{ \sqrt{3} } = \frac{1}{2} \sqrt{2} [/tex]
18. Cara menghitung sin 15 - cos 15 di bagi tan 15
Jawaban:
TrigonometriPenjelasan dengan langkah-langkah:
sin 15 - cos 15
--------------------
tan 15
sin (90-45) - cos (90-45)
-----------------------------------
tan (90-45)
cos 45 - sin 45
---------------------
cot 45
1/2√(2) - 1/2√(2)
-----------------------
1
0
---
1
0
Demikian
Semogamembantudanbermanfaat!
19. nilai dari sin 15 - cos 15/ tan 15 adalah
sin 15° - cos 15°/tan15°
~> sin 15° = sin (45° - 30°)
= sin45.cos30 - sin30.cos45
= √2/2.√3/2 - 1/2.√2/2
= √6/4 - √2/4 ~ (√6 - √2)/4 ✓
~> cos 15° = cos (45° - 30°)
= cos45.cos30 + sin45.sin30
= √2/2.√3/2 + √2/2.1/2
= √6/4 + √2/4 ~ (√6 + √2)/4 ✓
~> tan 15° = tan (45° - 30°)
= (tan45 - tan30)/1 + tan45.tan30
= (1 - 1/√3)/1 + 1/√3
= (√3 - 1)/√3/(√3 + 1)/√3
= (√3 - 1)/(√3 + 1) ✓
~> [(√6 - √2)/4 - (√6 + √2)/4]/[(√3 - 1)/(√3 + 1)]
-(√3 + 1)√2/2(√3 - 1)
(√6 + √2)/(2 - 2√3) ✓✓
20. Tolong jawabanya beserta caranya,, sin 15°- cos 15° / tan 15° =.....
[tex]\frac{\sin15-\cos15}{\tan15}\\\frac{\sin15-\cos15}{\frac{\sin15}{\cos15}}\\(\sin15-\cos15)(\frac{\cos15}{\sin15})\\\cos15-\frac{\cos^215}{\sin15}\\\cos(45-30)-\frac{(\cos(45-30))^2}{\sin(45-30)}\\(\cos45\cos30-\sin45\sin30)-\frac{(\cos45\cos30-\sin45\sin30)^2}{\sin45\cos30-\cos45\sin30}[/tex]
[tex](\frac{1}{4}\sqrt6+\frac{1}{4}\sqrt2)-\frac{(\frac{1}{4}\sqrt6+\frac{1}{4}\sqrt2)^2}{\frac{1}{4}\sqrt6-\frac{1}{4}\sqrt2}\\(\frac{\sqrt6+\sqrt2}{4})-\frac{(\frac{\sqrt6+\sqrt2}{4})^2}{\frac{\sqrt6-\sqrt2}{4}}\\\frac{\sqrt6+\sqrt2}{4}-\frac{\frac{6+2}{16}}{\frac{\sqrt6-\sqrt2}{4}}\\\frac{\sqrt6+\sqrt2}{4}-\frac{\frac{8}{16}}{\frac{\sqrt6-\sqrt2}{4}}\\\frac{\sqrt6+\sqrt2}{4}-\frac{1}{2}\times\frac{4}{\sqrt6-\sqrt2}\\\frac{\sqrt6+\sqrt2}{4}-\frac{2}{\sqrt6-\sqrt2}[/tex]
[tex]\frac{((\sqrt6)^2-(\sqrt2)^2)-8}{4(\sqrt6-\sqrt2)}\\\frac{(6-2)-8}{4(\sqrt6-\sqrt2)}\\\frac{4-8}{4(\sqrt6-\sqrt2)}\\\frac{-4}{4(\sqrt6-\sqrt2)}\\-\frac{1}{\sqrt6-\sqrt2}\times\frac{\sqrt6+\sqrt2}{\sqrt6+\sqrt2}\\-\frac{\sqrt6+\sqrt2}{6-2}\\\boxed{-\frac{\sqrt6+\sqrt2}{4}}[/tex]
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