document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1204248>Question 1204248</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given that cos 𝛼 =3/√10 and sin 𝛽 = −5/13, where −90° < 𝛼 < 0° and 180° < 𝛽 < 270°.<BR>\n" );
document.write( "Without using calculators,<BR>\n" );
document.write( "(a) find the values of sec(2𝛼 + 𝛽) and tan(2𝛼 + 𝛽);<BR>\n" );
document.write( "(b) determine the quadrant which contains 2𝛼 + 𝛽. Give reasons </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #840383 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=840383\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "I'll do the first portion of part (a) to get you started.\r<BR>\n" );
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document.write( "I'll use letters A and B in place of alpha and beta. \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28A%29+=+3%2Fsqrt%2810%29+=+3%2Asqrt%2810%29%2F10 ALIGN=MIDDLE  ALT=\"cos%28A%29+=+3%2Fsqrt%2810%29+=+3%2Asqrt%2810%29%2F10\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28B%29+=+-5%2F13 ALIGN=MIDDLE  ALT=\"sin%28B%29+=+-5%2F13\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-90+%3C+A+%3C+0 ALIGN=MIDDLE  ALT=\"-90+%3C+A+%3C+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=180+%3C+B+%3C+270 ALIGN=MIDDLE  ALT=\"180+%3C+B+%3C+270\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Use the pythagorean trig identity <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%5E2%28x%29+%2B+cos%5E2%28x%29+=+1 ALIGN=MIDDLE  ALT=\"sin%5E2%28x%29+%2B+cos%5E2%28x%29+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" > to go from <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28A%29+=+3%2Asqrt%2810%29%2F10 ALIGN=MIDDLE  ALT=\"cos%28A%29+=+3%2Asqrt%2810%29%2F10\"   DISABLED_style= \"max-width:90%; height:auto;\" > to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28A%29+=+-sqrt%2810%29%2F10 ALIGN=MIDDLE  ALT=\"sin%28A%29+=+-sqrt%2810%29%2F10\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28A%29+=+sqrt%2810%29%2F10 ALIGN=MIDDLE  ALT=\"sin%28A%29+=+sqrt%2810%29%2F10\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "I'll leave the scratch work steps for the student to do.\r<BR>\n" );
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document.write( "Which sine do we go for? The angle A is between -90 and 0, i.e. between 270 and 360, which places the angle in quadrant 4. This is where sine is negative.<BR>\n" );
document.write( "Therefore, we go with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28A%29+=+-sqrt%2810%29%2F10 ALIGN=MIDDLE  ALT=\"sin%28A%29+=+-sqrt%2810%29%2F10\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "If <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28B%29+=+-5%2F13 ALIGN=MIDDLE  ALT=\"sin%28B%29+=+-5%2F13\"   DISABLED_style= \"max-width:90%; height:auto;\" > then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28B%29+=+12%2F13 ALIGN=MIDDLE  ALT=\"cos%28B%29+=+12%2F13\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28B%29+=+-12%2F13 ALIGN=MIDDLE  ALT=\"cos%28B%29+=+-12%2F13\"   DISABLED_style= \"max-width:90%; height:auto;\" > when using the pythagorean trig identity. <BR>\n" );
document.write( "Angle B is in quadrant 3 where cosine is negative, which means we'll go for <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28B%29+=+-12%2F13 ALIGN=MIDDLE  ALT=\"cos%28B%29+=+-12%2F13\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Here's a short recap so far<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28A%29+=+-sqrt%2810%29%2F10 ALIGN=MIDDLE  ALT=\"sin%28A%29+=+-sqrt%2810%29%2F10\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sin%28B%29+=+-5%2F13 ALIGN=MIDDLE  ALT=\"sin%28B%29+=+-5%2F13\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%28B%29+=+-12%2F13 ALIGN=MIDDLE  ALT=\"cos%28B%29+=+-12%2F13\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Then,<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%282A%2BB%29+=+cos%282A%29cos%28B%29+-+sin%282A%29sin%28B%29 ALIGN=MIDDLE  ALT=\"cos%282A%2BB%29+=+cos%282A%29cos%28B%29+-+sin%282A%29sin%28B%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%282A%2BB%29+=+%282%2Acos%5E2%28A%29-1%29cos%28B%29+-+2%2Asin%28A%29cos%28A%29sin%28B%29 ALIGN=MIDDLE  ALT=\"cos%282A%2BB%29+=+%282%2Acos%5E2%28A%29-1%29cos%28B%29+-+2%2Asin%28A%29cos%28A%29sin%28B%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%282A%2BB%29+=+%282%2A%283%2Fsqrt%2810%29%29%5E2-1%29%2A%28-12%2F13%29+-+2%2A%28-sqrt%2810%29%2F10%29%2A%283%2Fsqrt%2810%29%29%2A%28-5%2F13%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=cos%282A%2BB%29+=+-63%2F65 ALIGN=MIDDLE  ALT=\"cos%282A%2BB%29+=+-63%2F65\"   DISABLED_style= \"max-width:90%; height:auto;\" > I skipped a few steps to get to this point. I'll let the student fill in the gaps.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sec%282A%2BB%29+=+1%2F%28cos%282A%2BB%29%29 ALIGN=MIDDLE  ALT=\"sec%282A%2BB%29+=+1%2F%28cos%282A%2BB%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sec%282A%2BB%29+=+-65%2F63 ALIGN=MIDDLE  ALT=\"sec%282A%2BB%29+=+-65%2F63\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "--------------\r<BR>\n" );
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document.write( "How can we help to confirm this?\r<BR>\n" );
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document.write( "cos(A) = 3/sqrt(10) leads to either <BR>\n" );
document.write( "A = arccos(3/sqrt(10)) = 18.434948822922<BR>\n" );
document.write( "or<BR>\n" );
document.write( "A = -arccos(3/sqrt(10)) = -18.434948822922\r<BR>\n" );
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document.write( "We go with the negative value to place angle A in quadrant 4.\r<BR>\n" );
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document.write( "Make sure that your calculator is set to degree mode.<BR>\n" );
document.write( "Then add on 360 to get a coterminal angle between 0 and 360.<BR>\n" );
document.write( "-18.434948822922 + 360 = 341.565051177078<BR>\n" );
document.write( "Notice how that value satisfies the interval 270 < A < 360.\r<BR>\n" );
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document.write( "sin(B) = -5/13 leads to <BR>\n" );
document.write( "B = arcsin(-5/13) = -22.6198649480404<BR>\n" );
document.write( "or<BR>\n" );
document.write( "B = 180-arcsin(-5/13) = 202.61986494804<BR>\n" );
document.write( "We go with the second result to place angle B in quadrant 3.\r<BR>\n" );
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document.write( "We have these approximate angle values<BR>\n" );
document.write( "A = 341.565051177078<BR>\n" );
document.write( "B = 202.61986494804\r<BR>\n" );
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document.write( "Then we can type this into a calculator:<BR>\n" );
document.write( "1/(cos(2*341.565051177078+202.61986494804)) - (-65/63)\r<BR>\n" );
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document.write( "The 1/(cos(2*341.565051177078+202.61986494804)) portion is us computing 1/(cos(2A+B)), i.e. sec(2A+B).<BR>\n" );
document.write( "We then subtract off (-65/33) to finish off the calculation.<BR>\n" );
document.write( "I'm using the idea that if x = y, then x-y = 0.<BR>\n" );
document.write( "So for example, 2+3 = 5 leads to 2+3-5 = 0. For such trivial examples like this, we won't need to use this rule. But it's useful for more complicated algebraic and trig expressions.\r<BR>\n" );
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document.write( "The result of this calculation should be very close to zero. It won't be zero itself because those A,B values we found were approximate. \r<BR>\n" );
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document.write( "Despite the difference of those values not being zero itself, it's quite possible your calculator rounds the result to 0.\r<BR>\n" );
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document.write( "My calculator shows the difference is the very small value of -1.998401 * 10^(-15) which is the same as writing <BR>\n" );
document.write( "<PRE STYLE=\"white-space: pre-wrap; white-space: -moz-pre-wrap; white-space: -pre-wrap; white-space: -o-pre-wrap; word-wrap: break-word;\">-0.000 000 000 000 001 998 401</PRE><BR>\n" );
document.write( "I have separated the decimal digits with a space every 3 digits to help make the number more readable. There are 14 zeros between the decimal point and the first copy of \"1\".\r<BR>\n" );
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document.write( "This all but confirms that we have the correct value of sec(2A+B).\r<BR>\n" );
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document.write( "Answer: sec(2A+B) = <font color=red size=4>-65/63</font><BR>\n" );
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