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ตรีโกณมิติของ ... - mwit.ac.th

6 o90o0< < r sin= cos= tan= csc= sec= cot=

ตัวอย างที่ 1 การหาค าฟ งก ชันตรีโกณมิติ ให ใช รูปสามเหล ี่ยมมุมฉากร ูป 3.2 หาค าฟ งก ชันตรีโกณมิติทั้งหก

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Transcription of ตรีโกณมิติของ ... - mwit.ac.th

1 6 o90o0< < r sin= cos= tan= csc= sec= cot= 1 22243x+= 2243x+= 25= 5=

2 54 sin== 45 csc== 53 cos== 35 sec== 34 tan== 43 cot== 1 2 45o sin 45o, cos 45o tan 45o 45o 1 = x 4 3 1 1 2 45o 45o 2 2221 o45sin=== 2221 o45cos=== 122 o45tan=== 3 30O 60O sin 30o, cos 30o.

3 Sin 60o cos 60o 2 3 60o 3 1 2 23 o60sin== 21 o60cos== 30o 1 3 2 21 o30sin== 23 o30cos== 30o 60o 1 1 2 2 30o 60o 1 1 2 2 3 (cofunctions of complementary angles are equal) = cos)o90(sin = sin)o90(cos = cot)o90(tan = tan)o90(cot = csc)o90(sec = sec)

4 O90(csc 216sino30sin= = 236coso30cos= = 336tano30tan= = 224sino45sin= = 224coso45cos= = 14tano45tan= = 233sino60sin= = 213coso60cos= = 33tano60tan= = 2sin 2)(sin 2cos 2)(cos 4 sin = (a) cos (b) tan (a) cos Pythagorean Identity 12cos2sin= + 12cos2) (= + ) (12cos= = (Reciprocal Identities) = csc1sin = sec1cos = cot1tan = sin1csc = cos1sec = tan1cot (Quotient Identities) = cossintan = sincoscot ( Pythagorean Identities) 12cos2sin= + =+ 2sec12tan = +2csc2cot1 (b) sin cos tan (Quotient Identities))

5 = cossintan = 5 tan = 5 (a) cot (b) sec (a) 51tan1cot= = (b) sec2 = 1 + tan2 sec2 = 1 + 52 sec2 = 26 sec = 26 (degree mode) cos 28O sec 28O cos 28O 28 sec 28O 28 sin 1 1 sin 1O 1 Enter ( cos cos ) x-1 Enter 6 sec(5O 40/ 12//) 5O 40/ 12// = ()()o360012o6040o5++ = sec sec(5O 40/ 12//)

6 = sec = 7 115 60O xh o60tan== h = x tan 60O h = 115 )3( (angle of elevation) (angle of depression)

7 H 60o x = 115 8 300 600 sin 21600300 sin=== = 30O 600 300